REESE    LIBRARY 

OF    THK 

UNIVERSITY   OF   CALIFORNIA, 

Received t^^&Zx^  /^  ^ 

Accessions  No..^f^f^if.S?     Shelf  No.... 


THE 


NATIONAL  'ARITHMETIC, 


ON   THE 


INDUCTIVE    SYSTEM; 


COMBINING    THE 


ANALYTIC  AND  SYNTHETIC  METHODS, 

TOGETHER  WITH  THE  CANCELLING  SYSTEM; 

FORMING    A. 

COMPLETE   MERCANTILE   ARITHMETIC. 


BY  BENJAMIN  GREENLEAF,  A.  M. 

PRINCIPAL  OP  BRADFORD   TEACHERS'   SEMINARY. 


tteto 

REVISED,   ENLARGED,    AND    MUCH    IMPROVED. 

BOSTON: 

ROBERT  S.  DAVIS,  AND   GOULD,  KENDALL,   &  LINCOLN. 

NEW  YORK:   PRATT,  WOODFORD,  &  Co.,  AND  CADY  &  BURGESS. 

PHILADELPHIA  :    THOMAS,  COWPERTHWAIT,  &  Co. 

BALTIMORE  :    GUSHING  &  BROTHER. 

And  sold  by  the  trade  generally. 

1848. 


Entered  according  to  Act  of  Coiigre§s,  in  th«  year  1847. 

BY  BENJAMIN  GKEENLEAF, 
In  the  Clerk'*  Office  of  the  District  Court  of  the  District  of  Massachusetts. 


STANDARD  BOSTON  SCHOOL  BOOKS. 


GREENLEAP'S  SERIES  OP  ARITHMETICS. 

1.  MENTAL  ARITHMETIC,  upon  the  Inductive  1'lan,  designed  for  beginners.    By 
Benjamin  Greenleaf,  A.  M.,  Principal  of  Bradford  (Mass.)  Teachers'  .Seminary. 

2.  INTRODrCTl6N  TO  THE  NATIONAL  ARITHMETIC,  designed  lor  Common 
Schools.    Twelfth  Improved  stereotype  edition. 

3.  THE  NATIONAL  ARITHMETIC,  for  advanced  Scholars  in  Common  Schools 
and  Academies.    Eighteenth  improved  stereotype  edition. 

COMPLETE  KEYS  TO  THE  INTRODUCTION  AND  NATIONAL  ARITHMETICS, 
containing  Solutions  and  Explanations,  for  Teachers  only.  (In  separate  volumes. ) 

%*  The  attentlbn  of  Teachers  and  Superintendents  of  Schools  generally  is  respect 
fully  Invited  to  this  popular  system  of  Arithmetic,  which  is  well  adapted  to  a^classes 
of  students.  The  'National  Arithmetic'  has  been  extensively  introduced  in  various 
sections  of  the  United  States,  and  is  highlv  recommended  by  many  distinguished 
teachers  who  have  used  it,  for  its  adaptedncss  to  give  students  a  thorough  practical 
knowledge  of  the  science.  It  is  the  text-book  in  the  Normal  Schools  in  Massachusetts, 
and  New  York  city,  and  the  best  schools  in  Boston,  New  York,  Philadelphia,  Rich 
mond,  Charleston,  .Savannah,  Mobile,  New  Orleans,  and  other  cities. 

PARKER'S  PROGRESSIVE  EXERCISES 

In  English  Composition      Forty -fifth  improved  stereotype  edition.    Price  34  cents. 

THE  CLASSICAL  READER, 

A  Selection  of  Lessons  in  Prose  and  Verse,  from  the  most  esteemed  English  and 
American  writers.  Intended  for  the  use  of  the  higher  classes  in  public  and  private 
seminaries.  By  Rev.  F.  W.  P.  Greenwood,  D.  D.  and  G.  B.  Emerson,  A.  M.  of  Bos- 
ton. Tenth  edition,  stereotyped.  With  an  engraved  Frontispiece. 

SMITH'S  CLASS  BOOS  OP  ANATOMY, 

Explanatory  oi  the  nrst  principles  of  Human  Organization  as  the  basis  of  Physical 
Education  ;  with  numerous  Illustrations,  a  full  Glossary,  or  explanation  of  Technical 
Terms,  and  practical  Questions  at  the  bottom  of  the  page.  Designed  for  Schools  and 
Families.  Ninth  stereotype  edition,  revised  and  enlarged. 

A  GRAMMAR  OP  THE  GREEK  LANGUAGE. 

By  Benjamin  Franklin  Kisk.  Twenty-sixth  improved  stereotype  edition.  Fist't 
Greek  Grammar  is  used  in  Harvard  University,  and  in  many  oth^r  distinguished  Col- 
legiate and  Academic  Institutions,  in  various  parts  of  the  United  Slates. 

FISK'S  GREEK  EXERCISES  (New  Edition.) 

Greek  Exercises;  containing  the  substance  of  the  Greek  Syntax,  illustrated  by 
passages  from  the  best  Greek  authors,  to  be  written  out  from  the  words  given  in  their 
simplest  form,  by  Benjamin  Franklin  Fisk.  '  Consuetudo  et  exercitutio  facilitatem 
niaxime  parit.'—  Quintil.  Adapted  to  the  Author's  'Greek  Grammar.'  Sixteenth 
stereotype  edition. 

Fist's  '  Greek  Exercises  '  are  «v>/{  adapted  to  illustrate  the  rules  of  the  Grammar, 
and  constitute  a  very  useful-accompaniment  thereto. 

LEVERETT'S   CJESAR'S  COMMENTARIES. 

Call  Julii  Csesaris  Commentaril  de  Bello  Gallico  ad  Codices  Parisinos  recensiti,  a 
N.  L.  Achaintre  et  N.  E.  Lfinaire.  Accesserimt  N'  tula:  Anglicse,  atque  Index  Uisto- 
ricus  et  Geographicus.  Curavit  F  P.  Leverett,  A.  M. 

FOLSOM'S  CICERO'S  ORATIONS. 

M.  T.  Ciceronis  Orationes  Qiuudain  Stlectae,  Notis  Illustrate.  [By  Charles  Fol- 
som,  A.  M.]  In  Usuin  Academiaj  Exoniensis.  Editio  stereotypa,  Tabulis  Analytic!* 
instructa. 

•  I  have  examined  with  some  attention  Csesar's  Commentaries,  edited  by  Leverett, 
and  Cicero's  Orations,  edited  by  Folsom,  and  am  happy  to  recommend  them  to  clas- 
sical teachers,  as  being,  In  my  estimation,  far  superior  to  any  other  editions  of  those 
works  to  which  students  in  this  country  have  general  access. 

'JOHN  J.  OWKN, 
Editor  of  'Xenophon't  Anabasis,'  and  Principal  of  Cornelius  Institute,  A.  Y.  City. 

ALGER'S   MURRAY'S    GRAMMAR,    AND    EXERCISES. 
ALGERB  MURRAY'S  READER  AND  INTRODUCTION. 

Published  by  ROBERT  S.  DA  VIS,  School- Book  Publisher,  BOSTON, 
and  fold  by  all  the  principal  Booksellers  throughout  the  United  State*. 

G^"  Also  constantly  for  sale,,  (hi  addition  to  his  mun  publications,)  a 
complete  assortment  of  School  Books  and  Stationery,  which  are  offered  to 
Booksellers,  Scliool  Committees,  and  Teachers,  on  very  liberal  terms. 


PREFACE. 


THE  National  Arithmetic  has  now  been  before  the  public  for 
nearly  twelve  years,  and  has  met  with  an  acceptance  far  be- 
yond the  original  expectation  of  the  author.  The  demand  for 
it  has  constantly  increased,  and  such  has  been  the  encourage- 
ment which  both  the  author  and  the  publisher  have  received 
from  teachers  of  the  highest  character,  and  from  the  public  gen- 
erally, that  the  work  has  been  thoroughly  revised  and  very  con- 
siderably enlarged,  particularly  in  the  department  of  demon- 
stration, and  is  now  presented  in  a  form  which,  it  is  believed, 
will  greatly  increase  its  value.  In  the  work  of  revision  and 
enlargement,  the  author  has  availed  himself  of  important  sug- 
gestions from  many  practical  teachers,  and  has  had  the  direct 
assistance  of  gentlemen  intimately  acquainted,  not  only  with 
the  business  of  teaching  Arithmetic,  but  also  with  the  higher 
branches  of  Mathematics.  His  own  labors  in  this  work  have 
been  hardly  less  than  in  the  original  preparation  of  the  book, 
and  he  is  confident  that  the  improvements  introduced  into  the 
present  edition  will  be  seen  and  appreciated  by  all  who  may 
compare  it  with  preceding  editions. 

It  has  been  the  author's  privilege,  for  more  than  thirty  years, 
to  be  engaged  in  the  business  of  instruction.  He  has  been 
acquainted  with  the  methods  of  communicating  knowledge 
which  were  formerly  practised,  and  has  endeavoured  to  make 
himself  familiar  with  all  the  improvements  in  this  respect 
which  distinguish  the  present  age  from  the  past.  The  present 
work  is  offered  to  the  public,  as  one  constructed  on  a  plan 
which  appears  to  the  author  better  adapted  to  meet  the  wants 
of  the  times  than  any  other  now  in  use.  The  end  to  be  sought 
in  the  study  of  Arithmetic  he  regards  as  twofold,  —  a  prac- 
tical knowledge  of  numbers  and  the  art  of  calculation,  and 
the  discipline  of  the  mental  powers  ;  and  the  present  work,  it 
is  believed,  will  be  found  fitted  to  these  two  objects.  It  is  in- 
tended to  be  comprehensive  in  its  principles,  and  sufficiently 
extensive  in  its  details ;  and  while  the  road  to  a  knowledge  of 


4  PREFACE. 

the  science  is  not  designed  to  be  unnecessarily  steep  and  rug- 
ged, the  author  does  not  desire  to  relieve  the  learner  of  all 
occasion  for  effort,  nor  make  him  feel  that  the  "  Hill  of  Sci- 
ence "  is  no  hill  at  all,  but  only  a  fiction  of  former  ages.  The 
author's  idea  is,  that,  in  order  to  become  a  thorough  and  accom- 
plished arithmetician,  one  must  study,  and  the  National  Arith- 
metic proposes  no  substitute  for  mental  exertion.  Still,  it  is 
not  designed  to  be  difficult  beyond  the  necessities  of  the  case, 
and  no  pupil,  who  is  faithful  to  himself,  will,  it  is  thought,  find 
reason  to  complain  that  enough  is  not  done  by  way  of  suitable 
illustration  to  facilitate  his  progress. 

It  is  the  opinion  of  some  teachers,  that  no  rules  should 
be  furnished  the  pupil  to  aid  him  in  performing  arithmetical 
questions,  but  that  every  pupil  should  form  his  own  rules  by 
the  process  of  induction.  But  the  author's  experience  has  led 
him  to  a  different  conclusion,  nor  does  he  think  that  the  inser- 
tion of  proper  rules,  in  a  work  like  the  present,  interferes  in  the 
least  with  the  necessity  of  study,  or  a  thorough  knowledge  of 
the  different  numerical  processes. 

The  National  Arithmetic  is  intended  to  be  complete  hi  itself; 
but  the  smaller  works  of  the  author  will  prepare  the  pupil  for 
an  easy  entrance  upon  the  study  of  it.  The  learner  can  omit 
the  more  difficult  parts  of  the  present  work  until  he  reviews  it, 
if  thought  advisable  by  the  teacher. 

A  few  rules,  which  are  omitted  in  some  works  on  Arithmetic 
at  the  present  day,  the  author  has  thought  best  to  retain,  — 
such  as  Practice,  Progression,  Position,  Permutation,  &c.  For, 
though  these  rules  may  not  in  themselves  be  of  great  practical 
utility,  yet,  as  they  are  well  adapted  to  improve  the  reasoning 
powers,  and  give  interest  to  the  higher  departments  of  arith- 
metical science,  it  is  deemed  desirable  to  place  them  within 
reach  of  the  student. 

In  closing  these  prefatory  remarks,  the  author  would  earnest- 
ly recommend  that  the  pupil  be  required  to  give  a  minute  and 
thorough  analysis  of  every  question  he  performs,  at  least  until 
he  has  proved  himself  familiar  with  all  the  principles  involved 
in  the  rule  under  consideration,  and  also  the  manner  of  their 
application.  He  would  further  recommend  a  frequent  and 
thorough  review  of  the  parts  of  the  work  which  the  pupil  has 
gone  over,  the  exercise  having  respect  mainly  to  the  principles 
involved  in  the  preceding  rules  and  examples, 

Bradford  Seminary,  September  1,  1847. 


CONTENTS. 


PAGES 

SECT.  INTRODUCTION 7-11 

1.  Numeration    .         .         .         .         .         .         .         .  13-17 

2.  Addition 18-21 

3.  Subtraction 21-24 

4.  Multiplicatior 24-29 

5.  Division 29-37 

6.  Contractions  in  Multiplication 37-39 

7.  Contractions  in  Division 39-41 

8.  Miscellaneous  Examples 41-43 

9.  Tables  of  Money,  Weights,  and  Measures          .         .  44  -  50 

10.  Compound  Addition 50-54 

11.  Compound  Subtraction     .         .         .         .         .         .         55-58 

Exercises  in  Compound  Addition  and  Subtraction  .         .58-60 

12.  Reduction 60-67 

13.  United  States  Money.     Addition,  Subtraction,  Multipli- 

cation, and  Division  of ;  Bills  in  .         .         .         .  67-77 

14.  Compound  Multiplication 78-81 

Bills  in  English  Money 81-83 

15.  Compound   Division    .         .         .         .         .         .  .84-88 

Questions  to  be  performed  by  Analysis      .         .         .  88-89 

16.  Vulgar  Fractions 89-109 

17.  Addition  of  Vulgar  Fractions 110-114 

18.  Subtraction  of  Vulgar  Fractions    .         .         .         .  114-121 

19.  Multiplication  of  Vulgar  Fractions     ....  121-125 

20.  Division  of  Vulgar  Fractions          ....  125-129 

21.  Questions  to  be  performed  by  Analysis     .         .         .  127-135 

22.  Decimal  Fractions.     Numeration  of  Decimal  Fractions  135  -  137 

23.  Addition  of  Decimals 137-138 

24.  Subtraction  of  Decimals 138-139 

25.  Multiplication  of  Decimals             .         .         .         .  139-141 

26.  Division  of  Decimals 141-142 

27.  Reduction  of  Decimals 142-145 

28.  Miscellaneous  Examples 145  -  147 

29.  Exchange  of  Currencies 149-152 

30.  Infinite  or  Circulating  Decimals       .         ,         ,         .  152-153 
Reduction  of  Circulating  Decimals         .         .         .  153-156 

31.  Addition  of  Circulating  Decimals      ....  156-157 

32.  Subtraction  of  Circulating  Decimals       .         .         .  157-158 

33.  Multiplication  of  Circulating  Decimals       .         .         .  158-159 

34.  Division  of  Circulating  Decimals  ....  159 

35.  Mental  Operations  in  Fractions,  &c.          .         .         .  159-161 

36.  Questions  to  be  performed  by  Analysis           .         .  162  -  164 

37.  Simple  Interest 164-172 

1* 


6  CONTENTS. 

38.  Partial  Payments 173-181 

39.  Miscellaneous  Problems  in  Inteiest         .         .         .  181  -  182 

40.  Compound  Interest 183-186 

41.  Discount 187-188 

42.  Percentage 188-189 

43.  Commission  and  Brokerage 189-191 

44.  Stocks 191-192 

45.  Insurance  and  Policies 192-193 

46.  Banking 193-194 

47.  Barter 195 

48.  Practice 196-198 

49.  Equation  of  Payments 199-201 

50.  Custom-House  Business 201  -  205 

51.  Ratio 205-207 

52.  Proportion       .         .         .  .         .         .         .207-217 

53.  Compound  Proportion,  or  Double  Rule  of  Three    .  217-221 

54.  Chain  Rule 221-223 

55.  Partnership,  or  Company  Business          .         .         .  223-225 

56.  Partnership  on  Time 225-227 

57.  General  Average 227-229 

58.  Profit  and  Loss 229-234 

59.  Duodecimals 234-238 

60.  Involution  ;    Evolution,   or  the   Extraction   of  Roots  ; 

Table  of  Powers 238-240 

61.  Extraction  of  the  Square  Root  .         .         .         .241-248 

62.  Extraction  of  the  Cube  Root          .         .         .         .  248  -  256 

63.  Arithmetical  Progression 257-261 

64.  Geometrical  Series,  or  Series  by  Quotient      .         .  261  -267 

65.  Infinite  Series 267-268 

66.  Discount  by  Compound  Interest     .... 

67.  Annuities  at  Compound  Interest         ....  269-272 

68.  Assessment  of  Taxes 272-275 

69.  Alligation 275-279 

70.  Permutations  and  Combinations     ....  279-282 

71.  Life  Insurance 282-285 

72.  Position 286-290 

73.  Exchange 290-305 

74.  Value  of  Gold  Coins 305-309 

75.  Geometry  (Definitions) 309-313 

Geometrical  Problems 313-316 

Mensuration  of  Solids  and  Superficies       .         .         .316-327 

76.  Gauging 327-328 

77.  Tonnage  of  Vessels 328  -  329 

78.  Mensuration  of  Lumber 330-331 

79.  Philosophical  Problems 331-335 

80.  Mechanical  Powers 335-340 

81.  Specific  Gravity 340-341 

82.  Strength  of  Materials  .         .         .         .         •         .  341-344 

83.  Astronomical  Problems 345-347 

84.  Miscellaneous  Questions 347-354 

APPENDIX.  —  Weights  and  Measures        .         .         .  355  -  360 


INTRODUCTION. 


HISTORY    OF    ARITHMETIC. 

THE  question,  who  was  the  inventor  of  Arithmetic,  or  in  what  age 
or  among  what  people  did  it  originate,  has  received  different  answers. 
In  ordinary  history  we  find  the  origin  of  the  science  attributed  by 
some  to  the  Greeks,  by  some  to  the  Chaldeans,  by  some  to  the  Phoe- 
nicians, by  Josephus  to  Abraham,  and  by  many  to  the  Egyptians. 
The  opinion,  however,  rendered  most  probable,  if  not  absolutely  cer- 
tain, by  modern  investigations  is,  that  Arithmetic,  properly  so  called, 
is  of  Indian  origin,  —  that  is,  that  the  science  received  its  first  defi- 
nite form  and  became  the  regular  germ  of  modern  Arithmetic  in  the 
regions  of  the  East. 

It  is  evident,  from  the  nature  of  the  case,  that  some  knowledge  of 
numbers  and  of  the  art  of  calculation  was  necessary  to  men  in  the  ear- 
liest periods  of  society,  since  without  this  they  could  not  have  per- 
formed the  simplest  business  transactions,  even  such  as  are  incidental 
to  an  almost  savage  state.  The  question,  therefore,  as  to  the  invention 
of  Arithmetic  deserves  to  be  considered  only  as  it  respects  the  origin 
of  the  science  as  we  now  have  it,  and  which,  as  all  scholars  admit, 
has  reached  a  surprising  degree  of  perfection.  And  in  this  sense  the 
honor  of  the  invention  must  be  awarded  to  the  Hindoos. 

The  history  of  the  various  methods  of  Notation,  or  the  different 
means  by  which  numbers  have  been  expressed  by  signs  or  characters, 
is  one  of  much  interest  to  the  advanced  and  curious  scholar,  but  the 
brevity  of  this  sketch  allows  us  barely  to  touch  upon  it  here.  Among 
the  ancient  nations  which  possessed  the  art  of  writing,  it  was  a  natu- 
ral and  common  device  to  employ  letters  to  denote  what  we  express 
by  our  numeral  figures.  Accordingly  we  find,  that,  with  the  Hebrews 
and  Greeks,  the  first  letter  of  their  respective  alphabets  was  used  for 
1,  the  second  for  2,  and  so  on  to  the  number  10,  —  the  latter,  however, 
inserting  one  new  character  to  denote  the  number  6,  and  evidently  in 
order  that  their  notation  might  coincide  with  that  of  the  Hebrew's, 
the  sixth  letter  of  the  Hebrew  alphabet  having  no  corresponding  one 
in  the  Greek. 

The  Romans,  as  is  well  known,  employed  the  letters  of  their  al- 
phabet as  numerals.  Thus,  I  denotes  1 ;  V,  5  ;  X,  10  ;  L,  50  ;  C, 
100  ;  D,  500 ;  and  M,  1000.  The  intermediate  numbers  were  ex- 
pressed by  a  repetition  of  these  letters  in  various  combinations ;  as 
II  for  2 ;  VI  for  6  ;  XV  for  15  ;  IV  for  4  ;  IX  for  9,  &c.  They  fre- 


8  INTRODUCTION. 

quently  expressed  any  number  of  thousands  by  the  letter  or  letters  de- 
noting so  many  units,  with  a  line  drawn  above  ;  thus,  V,  5,000 ;  VI, 
6,000  ;  X,  10,000 ;  T,  50,000  ;  Cf,  100,000 ;  M,  1,000,000. 

In  the  classification  of  numbers,  as  well  as  in  the  manner  of  ex- 
pressing them,  there  has  been  a  great  diversity  of  practice.  While 
we  adopt  the  decimal  scale  and  reckon  by  tens,  other  nations  have 
adopted  the  vicenary,  reckoning  by  twenties ;  others  the  quinary, 
reckoning  by  fives ;  and  others  the  binary,  reckoning  by  twos.  The 
adoption  of  one  or  another  of  these  scales  has  been  so  general,  that 
they  have  been  regarded  as  natural,  and  accounted  for  by  referring 
them  to  a  common  and  natural  cause.  The  reason  for  assuming  the 
binary  scale  probably  lay  in  the  use  of  the  two  hands,  which  were  em- 
ployed as  counters  in  computing  ;  that  for  employing  the  quinary,  in 
a  similar  use  of  the  five  fingers  on  either  hand ;  while  the  decimal  and 
vicenary  scales  had  respect,  the  former  to  the  ten  fingers  on  the  two 
hands,  and  the  latter  to  the  ten  fingers  combined  with  the  ten  toes  on 
the  naked  feet,  which  were  as  familiar  to  the  sight  of  a  rude,  uncivil- 
ized people  as  their  fingers.  —  It  is  an  interesting  circumstance  that 
in  the  common  name  of  our  numeral  figures,  digits  (digiti)  or  fingers, 
we  preserve  a  memento  of  the  reason  why  ten  characters  and  our 
present  decimal  scale  of  numeration  were  originally  adopted  to  express 
all  numbers,  even  of  the  highest  order. 

It  is  now  almost  universally  admitted  that  our  present  numeral  char- 
acters, and  the  method  of  estimating  their  value  in  a  tenfold  ratio  from 
right  to  left,  have  decided  advantages  over  all  other  systems,  both  of 
notation  and  numeration,  that  have  ever  been  adopted.  There  are 
those  who  think  that  a  duodecimal  scale,  and  the  use  of  twelve  numeral 
figures  instead  often,  would  afford  increased  facility  for  rapid  and  ex- 
tensive calculations,  but  most  mathematicians  are  satisfied  with  the 
present  number  of  numerals  and  the  scale  of  numeration  which  has 
attained  an  adoption  all  but  universal. 

It  was  long  supposed,  that  for  our  modern  Arithmetic  the  world  was 
indebted  to  the  Arabians.  But  this,  as  we  have  seen,  was  not  the 
case.  The  Hindoos  at  least  communicated  a  knowledge  of  it  to  the 
Arabians,  and,  as  we  are  not  able  to  trace  it  beyond  the  former  people, 
they  must  have  the  honor  of  its  invention.  They  do  not,  however, 
claim  this  honor,  but  refer  it  to  the  Divinity,  declaring  that  the  inven- 
tion of  nine  figures,  with  device  of  place,  is  to  be  ascribed  to  the  benefi- 
cent Creator  of  the  universe. 

But  though  the  invention  of  modern  Arithmetic  is  to  be  ascribed  to 
the  Hindoos,  the  honor  of  introducing  it  into  Europe  belongs  unques- 
tionably to  the  Arabians.  It  was  they  who  took  the  torch  from  the 
East  and  passed  it  along  to  the  West.  The  precise  period,  however, 
at  which  this  was  done,  it  is  not  easy  to  determine.  It  is  evident  that 
our  numeral  characters  and  our  method  of  computing  by  them  were  in 
common  use  among  the  Arabians  about  the  middle  of  the  tenth  cen- 
tury, and  it  is  probable  that  a  knowledge  of  them  was  soon  afterwards 
communicated  to  the  inhabitants  of  Spain  and  gradually  to  those  of  the 
other  European  countries.  Their  general  adoption  in  Europe  would 
not  seem  to  have  been  earlier  than  the  twelfth  or  thirteenth  century. 


HISTORY  OF  ARITHMETIC.  9 

The  science  of  Arithmetic,  like  all  other  sciences,  was  very  limited 
and  imperfect  at  the  beginning,  and  the  successive  steps  by  which  it 
has  reached  its  present  extension  and  perfection  have  been  taken  at 
long-  intervals  and  among  different  nations.  It  has  been  developed  by 
the  necessities  of  business,  by  the  strong  love  of  certain  minds  for 
mathematical  science  and  numerical  calculation,  and  by  the  call  for  its 
higher  offices  by  other  sciences,  especially  that  of  Astronomy.  In  its 
progress  we  find  that  the  Arabians  discovered  the  method  of  proof  by 
casting  out  the  9's,  and  that  the  Italians  early  adopted  the  practice  of 
separating  numbers  into  periods  of  six,  for  the  purpose  of  enumera- 
tion. To  facilitate  the  process  of  multiplication,  this  latter  people 
also  introduced,  probably  from  the  writings  of  Boethius,  the  long  neg- 
lected Table  of  Pythagoras. 

The  invention  of  the  Decimal  Fraction  was  a  great  step  in  the  ad- 
vancement of  arithmetical  science,  and  the  honor  of  it  has  generally 
been  given  to  Regiornontanus,  about  the  year  1464.  It  appears,  how- 
ever, more  properly  to  belong  to  Stevinus,  who  in  1582  wrote  an  ex- 
press treatise  on  the  subject.  The  credit  of  first  using  the  decimal 
point,  by  which  the  invention  became  permanently  available,  is  given 
by  Dr.  Peacock  to  Napier,  the  inventor  of  Logarithms ;  but  De  Mor- 
gan says  that  it  was  used  by  Richard  Witt  as  early  as  1613,  while  it 
is  not  shown  that  Napier  used  it  before  1617.  Circulating  Decimals 
received  but  little  attention  till  the  time  of  Dr.  Wallis,  the  author  of 
the  Arithmetic  of  Infinites.  Dr.  Wallis  died  at  Oxford,  in  1703. 

The  greatest  improvement  which  the  art  of  computation  ever  re- 
ceived was  the  invention  of  Logarithms,  the  honor  of  which  is  un- 
questionably due  to  Baron  Napier,  of  Scotland,  about  the  encT  of  the 
sixteenth  or  the  commencement  of  the  seventeenth  century. 

The  oldest  treatises  on  Arithmetic  now  known  are  the  7th,  8th,  9th, 
and  10th  books  of  Euclid's  Elements,  in  which  he  treats  of  proportion 
and  of  prime  and  composite  numbers.  These  books  are  not  contained 
in  the  common  editions  of  the  great  geometer,  but  are  found  in  the 
edition  by  Dr.  Barrow,  the  predecessor  of  Sir  Isaac  Newton  in  the 
mathematical  chair  at  Cambridge.  Euclid  flourished  about  300  B.  C. 

The  next  writer  on  Arithmetic  mentioned  in  history  is  Nicomachus, 
the  Pythagorean,  who  wrote  a  treatise  relating  chiefly  to  the  distinc- 
tions and  divisions  of  numbers  into  classes,  as  plain,  solid,  triangular, 
&c.  He  is  supposed  to  have  lived  near  the  Christian  era. 

The  next  writer  of  note  is  Boethius,  the  Roman,  who,  however, 
copied  most  of  his  work  from  Nicomachus.  He  lived  at  the  begin- 
ning of  the  sixth  century,  and  is  the  author  of  the  well-known  work  on 
the  Consolation  of  Philosophy. 

The  next  writer  of  eminence  on  the  subject  is  Jordanus,  of  Namur, 
who  wrote  a  treatise  about  the  year  1200,  which  was  published  by 
Joannes  Faber  Stapulensis  in  the  fifteenth  century,  soon  after  the 
invention  of  Printing. 

The  author  of  the  first  printed  treatise  on  Arithmetic  was  Pacioli, 
or,  as  he  is  more  frequently  called,  Lucas  de  Burgo,  an  Italian  monk, 
who  in  1484  published  his  great  work,  entitled  Summa  de  Arithmetica, 
&c.,  in  which  our  present  numerals  appear  under  very  nearly  their 
modern  form. 


10  INTRODUCTION. 

In  1522,  Bishop  Tonstall  published  a  work  on  the  Art  of  Computa- 
tion, in  the  Dedication  of  which  he  says  that  he  was  induced  to  study 
Arithmetic  to  protect  himself  from  the  frauds  of  money-changers  and 
stewards,  who  took  advantage  of  the  ignorance  of  their  employers.  In 
his  preparation  for  this  work,  he  professes  to  have  read  all  the  books 
which  had  been  published  on  this  subject,  adding,  also,  that  there  was 
hardly  any  nation  which  did  not  possess  such  books. 

About  the  year  1540,  Robert  Record.  Doctor  in  Physic,  printed  the 
first  edition  of  his  famous  Arithmetic,  which  was  afterward  augmented 
by  John  Dee,  and  subsequently  by  John  Mellis,  and  which  did  much  to 
advance  the  science  and  practice  of  Arithmetic  in  England  in  its  early 
stages.  This  work,  which  is  now  quite  a  curiosity,  effectually  de- 
stroys the  claim  to  originality  of  some  things  of  which  authors  much 
more  modern  have  obtained  the  credit.  In  it  we  find  the  celebrated 
case  of  a  will,  which  we  have  in  the  Miscellaneous  Questions  of 
Webber  and  Kinne,  and  which,  altered  in  language  and  the  time  of 
making  the  testament,  is  the  llth  Miscellaneous  Question  in  the  pres- 
ent work.  This  question  is,  by  his  own  confession,  older  than  Record, 
and  is  said  to  have  been  famous  since  the  days  of  Lucas  de  Burgo. 
In  Record  it  occurs  under  the  "  Rule  of  Fellowship."  Record  was 
the  author  of  the  first  treatise  on  Algebra  in  the  English  language. 

In  1556,  a  complete  work  on  Practical  Arithmetic  was  published  by 
Nicolas  Tartaglia,  an  Italian,  and  one  of  the  most  eminent  mathema- 
ticians of  his  time. 

From  the  time  of  Record  and  Tartaglia,  works  on  Arithmetic  have 
been  too  numerous  to  mention  in  an  ordinary  history  of  the  science. 
De  Morgan,  in  his  recent  work  (Arithmetical  Books),  has  given  the 
names  of  a  large  number,  with  brief  observations  upon  them,  and  to 
this  the  inquisitive  student  is  referred  for  further  information  in  regard 
both  to  writers  and  books  on  this  subject  since  the  invention  of  Print- 
ing. It  is  remarkable  that  De  Morgan  knew  next  to  nothing  of  any 
American  works  on  Arithmetic.  He  mentions  the  "  American  Ac- 
countant" by  William  Milns,  New  York,  1797,  and  gives  the  name 
of  Pike  (probably  Nicholas  Pike)  among  the  names  of  which  he  had 
heard  in  connection  with  the  subject.  He  had  also  seen  the  Memoir 
of  Zerah  Colburn.  Of  the  compilation  of  Webber  and  the  original 
works  of  Walsh  and  Warren  Colburn,  he  seems  to  have  been  entirely 
ignorant. 

The  various  signs  or  symbols,  which  are  now  so  generally  used  to 
abridge  arithmetical  as  well  as  algebraical  operations,  were  introduced 
gradually,  as  necessity  or  convenience  taught  their  importance.  The 
earliest  writer  on  Algebra  after  the  invention  of  Printing  was  Lucas 
de  Burgo,  above  mentioned,  and  he  uses  p  for  plus  and  m  for  minus, 
and  indicates  the  powers  by  the  first  two  letters,  in  which  he  was  fol- 
lowed by  several  of  his  successors.  After  this,  Steifel,  a  German, 
who  in  1544  published  a  work  entitled  Arithmetica  Integra,  added 
considerably  to  the  use  of  signs,  and,  according  to  Dr.  Hutton,  was  the 
first  who  employed  -f-  and  —  to  denote  addition  and  subtraction.  To 
denote  the  root  of  a  quantity  he  also  used  our  present  sign  ^/,  origi- 
nally r,  the  initial  of  the  word  radix,  root,  The  sign  =  to  denote 


HISTORY   OF  ARITHMETIC.  11 

equality  was  introduced  by  Record,  the  above-named  English  mathe- 
matician, and  for  this  reason,  as  he  says,  that  "  noe  2  thynges  can  be 
moar  equalle,"  namely,  than  two  parallel  lines.  It  is  a  curious  cir- 
cumstance that  this  same  symbol  was  first  used  to  denote  subtraction. 
It  was  also  employed  in  this  sense  by  Albert  Girarde,  who  lived  a 
little  later  than  Record.  Girarde  dispensed  with  the  vinculum  em- 
ployed by  Steifel,  as  in  3-J-4,  and  substituted  the  parenthesis  (S-f-4), 
now  so  generally  adopted.  The  first  use  of  the  St.  Andrew's  cross, 
X  ,  to  signify  multiplication  is  attributed  to  William  Oughtred,  an 
Englishman,  who  in  1631  published  a  work  entitled  Clavis  Mathe- 
matiaR,  or  Key  of  Mathematics. 

It  was  intended  to  notice  several  other  works,  ancient  and  modern, 
but  the  length  to  which  this  sketch  has  already  extended  forbids  it. 
We  must  not,  however, "omit  to  mention  two  American  works,  which 
have  done  much  for  the  cause  of  practical  Arithmetic  in  this  country. 
These  are  the  large  work  of  Nicholas  Pike,  first  published  about  1787, 
and  the  little  unpretending  "  First  Lessons  "  in  Arithmetic,  by  War- 
ren Colburn.  From  the  former  of  these  many  later  authors  have  bor- 
rowed much  that  is  useful,  and  the  latter  has  exerted  an  influence  on 
the  method  of  studying  Arithmetic  greater,  perhaps,  than  any  other 
modern  production.  No  better  elementary  work  than  that  of  Col- 
burn  has  ever,  it  is  believed,  appeared  in  any  language. 

We  had  thought  of  alluding  to  the  ancient  philosophic  Arithmetic, 
and  the  elevated  ideas  which  many  of  the  early  philosophers  had  of 
the  science  and  properties  of  numbers.  But  a  word  must  here  suffice. 
Arithmetic,  according  to  the  followers  of  Plato,  was  not  to  be  studied 
"  with  gross  and  vulgar  views,  but  in  such  a  manner  as  might  enable 
men  to  attain  to  the  contemplation  of  numbers  ;  not  for  the  purpose 
of  dealing  with  merchants  and  tavern-keepers,  but  for  the  improve- 
ment of  the  mind,  considering  it  as  the  path  which  leads  to  the  knowl- 
edge of  truth  and  reality."  These  transcendentalists  considered  per- 
fect numbers,  compared  with  those  which  are  deficient  or  superabun- 
dant, as  the  images  of  the  virtues,  which,  they  allege,  are  equally 
remote  from  excess  and  defect,  constituting  a  mean  between  them ; 
as  in  the  case  of  true  courage,  which,  they  say,  lies  midway  between 
audacity  and  cowardice,  and  of  liberality,  which  is  a  mean  between 
profusion  and  avarice.  In  other  respects,  also,  they  regard  this  anal- 
ogy as  remarkable:  perfect  numbers,  like  the  virtues,  are  "few  in 
number  and  generated  in  a  constant  order ;  while  superabundant  and 
deficient  numbers  are,  like  vices,  infinite  in  number,  disposable  in  no 
regular  series,  and  generated  according  to  no  certain  and  invariable 
law." 

We  conclude  this  brief  sketch  in  the  earnest  hope  that  the  noble 
science  of  numbers  may  ere  long  find  some  devoted  friend  who  shall 
collect,  arrange,  and  bring  within  the  reach  of  ordinary  students, 
much  more  fully  than  we  have  done,  the  scattered  details  of  its  long- 
neglected  history. 


ARITHMETICAL   SIGNS. 


—  Sign  of  equality ;  as   12  inches  =  1  foot  signifies  that  12 

inches  are  equal  to  one  foot. 

-|-  Sign  of  addition ;  as  8  -J-  6  =  14  signifies  that  8  added  to 
6  is  equal  to  14. 

—  Sign  of  subtraction ;  as  8  —  6  =  2,  that  is,  8  less  6  is  equal 

to  2. 
X   Sign  of  multiplication ;  as  7  X  6  —  42,  that  is,  7  multiplied 

by  6  is  equal  to  42. 
-j-  Sign  of  division ;  as  42  -i-  6  =  7,  that  is,  42  divided  by  6 

is  equal  to  7. 

-3^2-  Numbers  placed  in  this  manner  imply  that  the  upper  num- 
ber is  to  be  divided  by  the  lower  one. 
:  : :  :  Signs  of  proportion  ;  thus,  2  :  4  : :  6 :  12,  that  is,  2  has 

the  same  ratio  to  4  that  6  has  to  12 ;  and  such  numbers 

are  called  proportionals. 

12  —  3-|-4=z  13.  Numbers  placed  in  this  manner  show  that  3 

is  to  be  taken  from  12,  and  4  added  to  the  remainder. 

The  line  at  the  top  is  called  a  vinculum,  and  connects 

all  the  numbers  over  which  it  is  drawn. 
92   implies  that  9  is  to  be  raised  to  the  second  power ;  that  is, 

multiplied  by  itself. 
83   implies  that  8  is  to  be  multiplied  into  its  square,  or  to  be 

raised  to  the  third  power. 
\f  This  sign  prefixed  to  any  number  shows  that  the  square 

root  is  to  be  extracted. 
\/  This  sign  prefixed  to  a  number  shows  that  the  cube  root 

is  to  be  extracted. 
Sometimes  roots  are  designated  by  fractional  indices,  thus ; 

9*  denotes  the  square  root  of  9 ;  27*  denotes  the  cube 

root  of  27. 
(  )  [  ]  Parentheses  and  brackets  are  often  used  instead  of  a 

vinculum.     Thus,  (7  —  3)  x  5  =  60  -=-  3. 


03=  An  edition  of  this  work,  without  ansicers,  is  published  for  the  ac- 
commodation of  those  teachers  who  prefer  that  the  pupil  should  not  have 
access  to  them. 

A  KEY,  containing  solutions  and  explanations,  is  also  published  for  the 
convenience  of  teachers. 


ARITMTETIC, 


SECTION  I. 

ARITHMETIC  is  the  science  of  numbers,  and  the  art  of  com- 
puting by  them. 

The  operations  of  Arithmetic  are  performed  principally  by 
Addition,  Subtraction,  Multiplication,  and  Division. 


NUMERATION. 

NUMERATION  teaches  to  express  the  value  of  numbers,  either 
by  words  or  characters. 

Numbers  in  Arithmetic  are  expressed  by  the  ten  following 
characters,  which  are  called  numeral  figures  ;  viz.  1  (owe),  2 
(two],  3  (three),  4  (four),  5  (five),  6  (six),  7  (seven),  8 
(eight),  9  (nine),  Q  (cipher,  or  nothing). 

The  first  nine  of  these  figures  are  called  significant,  as  dis- 
tinguished from  the  cipher,  which  is  of  itself  insignificant. 

Besides  this  value  of  the  rfumerical  figures,  they  have  another 
value,  dependent  on  the  place  which  they  occupy,  when  con- 
nected together.  This  is  illustrated  by  the  following  table  and 
its  explanation. 

2 


14  NUMERATION.  [SECT.  i. 


I 


J.I  i       H       4 

"§  o 

S3  c 

a     i? 


£ 

2 

o 

i 

i 

U3 

| 

CO 

fl 

•d 

CO 

C 

1 

*-^ 

3 

QJ 

gH 

^3 

Q 

'3 

3 

ffi 

H 

H 

ffi 

H 

7 

6 

5 

4 

3 

2 

1 

8 

7 

6 

5 

4 

3 

2 

9 

8 

7 

6 

5 

4 

3 

9 

8 

7 

6 

5 

4 

9 

8 

7 

6 

5 

9 

8 

7 

6 

9 

8 

7 

9 

8 

9 

Here  any  figure  occupying  the  first  place,  reckoning  from 
right  to  left,  denotes  only  its  simple  value  or  number  of  units. 
But  the  figure  standing  in  the  second  place  denotes  ten  times 
its  simple  value  ;  that  occupying  the  third  place,  a  hundred 
times  its  simple  value,  and  so  on  to  any  required  number  of 
places  ;  the  value  of  any  figure  being  always  increased  tenfold 
by  its  removal  one  place  to  the  left. 

Thus,  in  the  number  1834,  the  4  in  the  first  place  denotes 
only  four  units,  or  simply  4  ;  the  3  in  the  second  place  signi- 
fies three  tens,  or  thirty  ;•  the  8  in  the  third  place  signifies  eighty 
tens,  or  eight  hundred  ;  and  the  1  in  the  fourth  place,  one  thou- 
sand ;  so  that  the  whole  number  is  read  thus,  —  one  thousand 
eight  hundred  thirty-four. 

Although  the  cipher  has  no  value  of  itself,  when  standing 
alone,  yet,  being  joined  to  the  right  hand  of  significant  figures, 
it  increases  their  value  in  a  tenfold  proportion ;  thus,  5  signifies 
simply  five,  while  50  denotes  five  tens,  or  fifty  ;  500,  five  hun- 
dred, and  so  on. 

NOTE. — The  idea  of  number  is  the  latest  and  most  difficult  to  form. 
Before  the  mind  can  arrive  at  such  an  abstract  conception,  it  must  be 
familiar  with  that  process  of  classification,  by  which  we  successively  re- 
mount from  individuals  to  species,  from  species  to  genera,  from  genera  to 
orders.  The  savage  is  lost  in  his  attempts  at  numeration,  and  significant- 
ly expresses  his  inability  to  proceed  by  holding  up  his  expanded  fingers 
or  pointing  to  the  hair  of  his  head.  See  Lacroix. 


SECT.    I.] 


NUMERATION. 


15 


NUMERATION  TABLE. 

The  following  is  the  French  method  of  enumeration,  and  is 
in  general  use  in  the  United  States  and  on  the  continent  of 
Europe. 

In  order  to  enumerate  any  number  of  figures  by  this  method, 
they  should  be  separated  by  commas  into  divisions  of  three 
figures  each,  as  in  the  annexed  table.  Each  division  will  be 
known  by  a  different  name.  The  first  three  figures,  reckon- 
ing from  right  to  left,  will  be  so  many  units,  tens,  and  hun- 
dreds, and  the  next  three  so  many  thousands,  and  the  next 
three  so  many  millions,  &c. 


g  Vigintillions. 
Jj  Novemdecillions. 
jg  Octodecillions. 
J§  Septendecillions. 
i  Sexdecillions. 
jl  Quindecillions. 
S  Quatuordecillions. 
£  Tridecillions. 
i  Duodecillions. 
J3  Undecillions. 
8  Decillions. 
i  Nonillions. 
i  Octillions. 
I  Septillions. 
%  Sextillions. 
jl  Quintillions. 
j8  Quadrillions. 
I  Trillions. 
E  Billions. 
I  Millions, 
jl  Thousands. 
JS  Units. 


The  value  of  the  numbers  in  the  an- 
nexed table,  expressed  in  words,  is 
One  hundred  twenty-three  vigintillions, 
four  hundred  fifty-six  novemdecillions, 
seven  hundred  eighty-nine  octodecil- 
lions,  one  hundred  twenty-three  septen- 
decillions,  four  hundred  fifty-six  sexde- 
cillions,  seven  hundred  eighty-nine  quin- 
decillions,  one  hundred  twenty-three 
quatuordecillions,  four  hundred  fifty-six 
tridecil lions,  seven  hundred  eighty-nine 
duodecillions,  one  hundred  twenty-three 
undecillions,  four  hundred  fifty-.six  de- 
cillions,  seven  hundred  eighty-nine  no- 
nillions,  one  hundred  twenty-three  oc- 
tillions, four  hundred  fifty-six  septil lions, 
seven  hundred  eighty-nine  sextillions, 
one  hundred  twenty-three  quintillions, 
four  hundred  fifty-six  quadrillions,  seven 
hundred  eighty-nine  trillions,  one  hun- 
dred twenty-three  billions,  four  hundred 
fifty-six  millions,  seven  hundred  eighty- 
nine  thousands,  one  hundred  twenty- 
three  units. 


16 


NUMERATION. 


[SECT. 


NUMERATION  TABLE.  The    following   is   the    old   English 

method  of  enumeration,  but  it  has  be- 
come almost  obsolete  in  this  country. 
In  order  to  enumerate  any  number  of 
figures  by  this  method,  they  should  be 
separated  by  semicolons  into  divisions 
of  six  figures  each,  and  each  division 
separated  in  the  middle  by  a  comma,  as 
in  the  annexed  table.  Each  division 
will  be  known  by  a  different  name. 
The  first  three  figures,  in  each  division, 
reckoning  from  right  to  left,  will  be  so 
so  many  units,  tens,  and  hundreds  of 
the  name  belonging  to  the  division,  and 
the  three  on  the  left  will  be  so  many 
thousands  of  the  same  name.  The 
value  of  the  numbers  in  the  annexed 
table,  expressed  in  words,  is  Three 
hundred  and  seventeen  thousand,  eight 
hundred  and  ninety-seven  tridecillions  ; 
four  hundred  and  thirty-one  thousand, 
thirty-two  duodecillions ;  six  hundred 
thirty-nine  thousand,  eight  hundred  six- 
ty-four undecillions  ;  three  hundred  six- 
ty-one thousand,  three  hundred  sixteen 
decillions  ;  four  hundred  sixty-one  thou- 
sand, three  hundred  fifteen  nonillions ; 
one  hundred  twenty-three  thousand,  six 
hundred  seventy-five  octillions;  eight 
hundred  sixteen  thousand,  one  hundred 
thirty-one  sepfillions ;  one  hundred  twen- 
ty-three thousand,  four  hundred  fifty-six 
sextillions;  one  hundred  twenty-three 
thousand,  six  hundred  fourteen  quintil- 
lions ;  three  hundred  fifteen  thousand, 

one  hundred  thirty-one  quadrillions;  three  hundred  ninety- 
eight  thousand,  eight  hundred  thirty-two  trillions;  five  hun- 
dred sixty -three  thousand,  eight  hundred  seventy-one  billions; 
three  hundred  fifty-one  thousand,  six  hundred  fifteen  millions ; 
one  hundred  twenty-three  thousand  five  hundred  sixty-one. 

NOTE.— The  student  must  be  familiar  with  the  names,  from  units  to 
tridecillions,  and  from  tridecillions  to  units,  so  that  he  may  repeat  them 
with  facility  either  way. 


S-  Thousands. 
|  Tridecillions. 
S»  Thousands. 
§  Duodecillions. 
1  Thousands. 
J  Undecillions. 
^  Thousands. 
5  Decillions. 
J2  Thousands. 
S  Nonillions. 
8  Thousands. 
Jl  Octillions. 
js  Thousands. 
js  Septillions. 
jg  Thousands. 
jg  Sextillions. 
jS  Thousands. 
£  Quintillions. 
j£  Thousands. 
js  Quadrillions. 
1  Thousands. 
J  Trillions. 

1  Thousands. 
"§  Billions. 

J2  Thousands, 
jj;  Millions. 
$  Thousands. 

2  Units. 


SECT,  i.]  NUMERATION.  17 

Let  the  following  numbers  be  written  in  words  :  — 

706 

313,461 

604,021 

3,607,005 

607,081,107 

470,803,020 

7,801,410,909 

322,172,517,101 

607,100,001,070 

407,000,010,703,801 

200,070,007,801,000 

670,812,000,170,063,891 

478,127,815,016,666,060,707 

800,800,800,800,800,800,800,800 

127,081,061,071,081,010,009,007,007 

407,144,140,070,060,700,007,101,800,808 

Let  the  following  numbers  be  written  in  figures  :  *  — 

1.  Twenty-nine. 

2.  Four  hundred  and  seven. 

3.  Twenty-three  thousand  and  seven. 

4.  Five  millions  and  twenty-seven. 

5.  Seven  millions,  two  hundred  five  thousand  and  five. 

6.  Two  billions,  two  hundred  seven  millions,  six  hundred 
four  thousand  and  nine. 

7.  One  hundred  five  billions,  nine  hundred  nine  millions, 
three  hundred  eight  thousand  two  hundred  and  one. 

8.  Nine  quintillions,  eight  billions  and  forty-six. 

9.  Fifteen  quintillions,  thirty-one  millions  and  seventeen. 

10.  Five  hundred  seven  septillions,  two  hundred  three  tril- 
lions, fifty-seven  millions  and  eighteen. 

11.  Nine  nonillions,  forty-seven  trillions,  seven  billions,  two 
millions,  three  hundred  ninety-two. 

12.  Fifteen  duodecillions,  ten  trillions,  one  hundred  twenty- 
seven  billions,  twenty-six  millions,  three  hundred  twenty  thou- 
sand four  hundred  twenty-six. 

*  To  express  numbers  by  figures,  begin  at  the  left  hand  with  the 
highest  order  mentioned,  and,  proceeding  to  units,  write  in  each  succes- 
sive order  the  figure  which  denotes  the  given  number  in  that  order. 
If  any  of  the  intervening  orders  are  not  mentioned  in  the  given  number, 
supply  their  places  with  ciphers. 
2* 


18  ADDITION.  [SECT.  H. 

SECTION  II. 
ADDITION. 

ADDITION  is  the  collecting  of  numbers  to  find  their  sum. 

1.  A  man  has  three  farms  ;  the  first  contains  378  acres,  the 
second  586  acres,  and  the  third  168  acres.  How  many  acres 
are  there  in  the  three  farms  ? 

In  this  question,  the  units  are  first  added,  and 
OPERATION,    their  sum  is  found  to  be  22 ;  in  22  units  there  are 
Acres.         two  tens  and  two  units.     The  two  units  are  writ- 
ten under  the  column  of  units,  and  the  2  (tens) 
586          are  carried  to  be  added  with  the  tens,  which  are 
found  to  amount  to  23  tens  —  2  hundreds  and 
1132          3  tens.     The  3  is  written  under  the  column  of 
tens,   and  the   2    (hundreds)  is  carried   to   the 
column  of  the  hundreds,  which  amount  to  11=1  thousand  1 
hundred.     The  whole  of  which  is  set  down.     Hence  the  pro- 
priety of  the  following 

RULE. 

Write  units  under  units,  tens  under  tens,  <SfC.  Then  begin  at  the 
bottom  and  add  the  units  upivards,  and,  if  the  amount  be  less  than  ten, 
set  it  down  under  the  column  of  units ;  but  if  the  amount  be  ten  or  more, 
write  down  the  unit  figure,  and  add  the  figure  denoting  the  number 
of  tens  to  the  column  of  tens.  Thus  proceed,  till  every  column  of 
figures  is  added,  writing  down  on  the  left  the  sum  total  of  the  left- 
hand  column,  and  the  result  will  be  the  sum  of  the  whole  as  required. 

PROOF. 

Begin  at  the  top  and  add  all  the  columns  downwards,  in  the 
same  manner  as  they  were  before  added  upwards  ;  then,  if  the 
two  sums  agree,  the  work  may  be  presumed  to  be  correct. 

2.  3.  4.  6.  6. 

Dollars.          Bushels.  Pecks.  Tons.  Miles. 

15  76  765  126  969 

26  48  381  384  872 

18  59  872  876  446 

91  81  315  243  392 

150  264  2333  1629  2679 


SECT.    II.] 


ADDITION. 


19 


8. 
Barrels. 

123 
456 
789 
341 

1709 

15. 

Ounces. 

7891 
3245 
6789 
1234 

5678 

24837 


9.  10.  11.  12.  13.  14. 

Pounds.  Acres.  Cents.  Eagles.  Rods.  Poles. 

678  456  789  456  781  889 

901  789  987  781  175  776 

278  127  123  197  564  432 

633  815  321  715  337  876 

2490  2187  2220  2149  1857  2973 


16. 
Inches. 

3256 

7890 
1234 
5678 
7801 

25859 


17. 
Rods. 

6789 
1234 
5678 
6543 
1234 

21478 


18. 
Furlongs. 

1234 

5678 
9012 
3456 
7891 

27271 


19. 

Cords. 

4567 
8912 
3456 
7891 
4567 

29393 


20. 
Feet. 

4561 
7890 
7658 

8888 
9199 

38196 


21. 

Hogsheads. 

1789 
6543 
2177 
8915 
6781 
4325 


27. 
Shillings. 

78956 
32167 
41328 
45678 
13853 
71667 


23 

Miles. 

7890 
1070 
4437 
6789 
5378 
1234 


24. 

Dollars. 
1785 

5678 
9137 
8171 
1888 
1919 


Acres. 

789516 
377895 
378567 
832156 
789567 
813138 


Tons. 

12345 
87655 
34517 
65483 
79061 
20939 


Miles. 

34567 
78901 
32199 

17188 
88888 
12345 


33. 

Roods. 

451237 
813715 
679919 
787651 
637171 
813785 


30. 

Trees. 

76717 
77777 
67890 
71444 
47474 
16175 


34. 

Poles. 

1234567 
8901234 
5678901 
3456789 
5432115 
7177444 


Pence. 

4371 

1699 
1098 
8816 
6171 
7185 

31. 

Loads. 

56789 
12345 
67819 
34567 

71888 
33197 


35. 

Yards. 

789123 
456789 
987654 
357913 
245678 
999999 


20 


ADDITION. 


[SECt.  II. 


36. 

123456 
789012 
345678 
901234 
567890 
987654 
321032 
765437 


37. 

876543 
789112 
345678 
965887 
445566 
788743 
399378 
456789 


33. 

789012 
345678 
901234 
789037 
891133 
477666 
557788 
888878 


39. 

987654 
456112 
222333 
456789 
987654 
321178 
123456 
789561 


42. 
Pounds. 

12004 

32 

1 

7836 

100 

46 

3 

6176 

32 

91876 


43. 
Ounces. 

30530 
31643 

26798 

28578 

34383 

29340 

283649 

300000 

264088 

357477 


40. 

678953 
467631 
117777 

888888 
444444 
667679 
998889 
671236 

44. 
Grains. 

276605 
3980839 
4183478 
31881050 
3837156 
4801393 
5067696 
5640426 
4344737 
1937678 


45.  Add  the  following  numbers,  763,  4663,  37,  49763,  6178, 
and  671.  Ans.  62075. 

46.  A  butcher  sold  to  A  369  Ibs.  of  beef,  to  B  169  Ibs.,  to  C 
861  Ibs.,  to  D  901  Ibs.,  to  E  71  Ibs.,  and  to  F  8716  Ibs. ;  what 
did  they  all  receive  ?  Ans.  1 1087  Ibs. 

47.  A  owes  to  one  creditor  596  dollars,  to  another  3961,  to 
another  581,  to  another  6116,  to  another  469,  to  another  506, 
to  another  69381,  and  to  another  1261.     What  does  he  owe 
them  all?  Ans.  $82871. 

48.  If  a  boy  earn  17  cents  a  day,  how  much  will  he  earn  in 
7  days  ?  Ans.  1 19  cts. 

49.  If  a  man's  wages  be  19  dollars  per  month,  what  are  they 
per  year  ?  Ans.  $  228. 

50.  If  a  boy  receive  a  present  every  New  Year's  day  of  1783 
dollars,  how  much  money  will  he  possess,  when  he  is  21  years 
old?  Ans.  $37443. 

51.  How  many  inhabitants  were  there  in  Essex  county,  Mass., 
in  1840,  Haverhill  having  4336,  Amesbury  2471,  Andover  5207, 
Beverly  4689,  Bradford   2222,  Boxford  942,  Danvers    5020, 
Essex  1450,  Georgetown  1540,  Gloucester  6:J50,  Hamilton  818, 


SECT.  HI.]  SUBTRACTION.  21 

Ipswich  3000,  Lynn  9369,  Lynnfield  707,  Manchester  1355, 
Marblehead  5575,  Methuen  2251,  Middleton  657,  Newbury 
3789,  Newburyport  7161,  Roekport  2650,  Rowley  1203,  Salem 
15082,  Salisbury  2739,  Saugus  1098,  Topsfield  1059,  Wenham 
689,  West  Newbury  1560?  Ans.  94,989. 

52.  How  many  were  the  members  of  Congress  in  1846, 
there  being  2  Senators  from  each  State,  and  Maine  sending  7 
Representatives,  New  Hampshire  4,  Massachusetts  10,  Rhode 
Island  2,  Connecticut  4,  Vermont  4,  New  York  34,  New  Jer- 
sey 5,  Pennsylvania  24,  Delaware  1,  Maryland  6,  Virginia  15, 
North  Carolina  9,  South  Carolina  7,  Georgia  8,  Alabama  7, 
Mississippi  4,  Louisiana  4,  Tennessee  11,  Kentucky  10,  Ohio 
21,  Indiana  10,  Illinois  7,  Missouri  5,  Arkansas  1,  Michigan  3, 
Florida  1,  Texas  2  ?  Ans.  282. 

53.  According  to  the  census  of  1840,  Maine  had  501,793  in- 
habitants,  New  Hampshire  284,574,  Massachusetts  737,699, 
Rhode  Island  108,830,  Connecticut  309,978,  Vermont  291,948 
New   York   2,428,921,   New   Jersey   373,306,    Pennsylvania 
1,724,033,   Delaware  78,085,   Maryland  469,232,  District  of 
Columbia  43,712,  Virginia  1,239,797,  North  Carolina  753,419, 
South  Carolina  594,398,  Georgia  691,392,  Kentucky  779,828, 
Tennessee  829,210,  Ohio  1,519,467,  Indiana  685,866,  Missis- 
sippi 375,651,  Missouri  383,702,  Illinois  476,183,  Louisiana 
352,411,    Alabama    590,756,    Michigan    212,267,    Arkansas 
97,574,  Florida  54,477,  Wisconsin  30,945,  Iowa  43,1 12,  and 
on  board  U.  S.  vessels  6,100.    What  was  the  whole  number  of 
inhabitants?  Ans.  17,068,666. 


SECTION  III. 

SUBTRACTION. 

SUBTRACTION  teaches  to  find  the  difference  between  two  num- 
bers by  taking  the  less   from  the  greater. 
OPERATION.          In  this  question,  we  take  3  units  from  5  units 
From     935     and  2  units  remain,  which  we  write  down  under 
Take     673     units,  as  the  first  figure  in  the  answer.     In  at- 
"ogo     tempting  to  take  the  7  tens  from  3  tens  we  find 
a  difficulty,  as  7  cannot  be  taken  from  3.     We 
therefore  borrow    1  (hundred)   from   the   9  (hundred),  which 
being  equal  to  10  tens,  we  add  it  to  the  3  tens  in  the  upper  line, 
making  13  tens ;  from  which  we  take  7  tens,  and  6  tens  re- 


22 


SUBTRACTION. 


[SECT.  in. 


main,  which  we  write  down  under  the  place  of  tens.  We 
then  proceed  to  the  hundreds.  As  we  have  borrowed  1  from 
the  9  hundreds,  the  9  is  too  large  by  1.  We  must  therefore 
take  the  6  (hundreds)  from  8  hundreds  and  there  will  re- 
main 2  (hundreds).  We  therefore  write  down  the  2  in  the 
place  of  hundreds.  Or,  because  the  9  is  too  large  by  1,  we 
may  add  1  to  the  6,  and  say  7  from  9  and  2  will  remain. 
Hence  the  following 

RULE. 

Place  the  less  number  under  the  greater ;  units  under  units,  tens  un- 
der tens,  <5fc.  Begin  with  the  units,  and  if  the  lower  figure  be  smaller 
than  the  one  above  it,  write  the  difference  below.  But,  if  the  upper  figure 
be  less  than  the  lower,  then  add  ten  to  the  upper  one,  and  write  the  dif- 
ference between  the  sum  thus  obtained  and  the  lower  figure.  Then  carry 
or  add  one  to  the  lower  figure  of  the  next  column,  and  proceed  as  before, 
till  all  the  numbers  are  subtracted,  and  the  result  will  be  the  difference. 

NOTE.  —  The  upper  number  is  called  the  Minuend,  from  the  Latin 
word  minuendum,  signifying  to  be  made  less ;  and  the  lower  one  the 
Subtrahend,  from  subtrakendum,  to  be  taken  away.  The  result  is  the 
Remainder. 

PROOF. 

Add  the  remainder  to  the  subtrahend,  and,  if  their  sum  be 
like  the  minuend,  the  work  may  be  considered  correct. 

6. 
Inches. 

11630078 
1919179 
9710899 


2. 

3. 

4. 

5. 

£. 

Cwt. 

Tons. 

Miles. 

Minuend, 

79 

86 

469 

876315123 

Subtrahend, 

24 

25 

183 

177897638 

55 

61 

286 

698417485 

From 
Take 


9. 

Minutes. 

6178 
1769 


10. 

Pecks. 

4567 
1978 


11. 

Barrels. 

From       765116 
Take       716669 

15. 

Hogsheads. 

From   611000 
Take   199999 


12. 

Degrees. 

56789 
10091 

16. 
Bushels. 

617853 
190909 


13. 

Furlongs. 

56781 
39109 

17. 
Yards. 

7111111 
909009 


14. 
Tons. 

71678 
18819 

18. 

Pounds. 

999000 
199919 


SECT.  III.] 


SUBTRACTION. 


23 


From 
Take 


19. 

Roods. 

100200 
98761 


23. 

Dollars. 

From  10000000 
Take   9099019 


20. 

Acres. 

511799 
419109 


21. 

Poles. 

610000 
166666 


Eagles. 

99999999 
1000919 


22. 
Cords. 

789111 
171670 

25. 
Guineas. 

888888 
99999 


26.  ' 

27. 

Seconds. 

Hours. 

From 

100200300400500 

600700800900 

Take 

90807060504039 

191818917185 

28.               29. 

30. 

Months.            Days. 

Weeks. 

From 

61567101      1000000 

10000000 

Take 

91678           1 

9999999 

31.  From 

6767851  take 

81715 

32.  From 

761619161  take 

916781 

33.  From 

31671675  take 

361784 

34.  From 

16781321  take 

100716 

35.  From 

1002007000  take 

5971621 

36.  From 

91611237  take 

6718538 

37.  From 

4637561  take 

4171135 

38.  From 

88895651  take 

3147618 

39.  From 

1111111  take 

99999 

40.  From 

7163878  take 

11001 

41.  From 

8887771  take 

81106 

42.  From 

1379156  take 

76716 

43.  From 

3671652  take 

36 

Sum  of  the  remainders,  2004466259. 

44.  Sir  Isaac  Newton  was  born  in  the  year  1642,  and  he  died 
in  1727 ;  how  old  was  he  at  the  time  of  his  decease  ? 

Ans.  85  years. 

45.  Gunpowder  was  invented  in  the  year  1330 ;  how  long 
was  this  before  the  invention  of  printing,  which  was  in  1441  ? 

Ans.  Ill  years. 

46.  The  mariner's  compass  was  invented  in  Europe  in  the 


24  MULTIPLICATION.  [SECT.  IT. 

year  1302  ;  how  long  was  this  before  the  discovery  of  America 
by  Columbus,  which  happened  in  1492  ?         Ans.  190  years. 

47.  What  number  is  that,  to  which  if  6956  be  added,  the 
sum  will 'be  one  million?  Ans.  993044. 

48.  A  man  bought  an  estate  for  seventeen  thousand  five  hun- 
dred and  sixty-five  dollars,  and  sold  it  for  twenty-nine  thousand 
three  hundred  and  seventy-five   dollars.     Did  he  gain  or  lose, 
and  how  much?  Ans.  Gained  $11810. 

49.  Bought  a  pair  of  oxen  for  85  dollars,  a  horse  for  126 
dollars,  three  cows  at  25  dollars  apiece ;  and  sold  the  whole 
for  three  hundred  dollars  ;  how  much  did  I  gain  ?     Ans.  $14. 

50.  Bonaparte  was  declared  emperor  in  1804 ;  how  many 
years  since  ? 

51.  The  union  of  the  government  of  England  and  Scotland 
was  in  the  year  1603 ;  how  long  was  it  from  this  period  to  the 
time  of  the  declaration  of  the  independence  of  the  United 
States  ?  Ans.   173  years. 

52.  Jerusalem  was  taken  and  destroyed  by  Titus  in  the  year 
70 ;  how  long  was  it  from  this  period  to  the  time  of  the  first 
Crusade,  which  was  in  the  year  1096  ?        Ans.  1026  years. 


SECTION  IV. 
MULTIPLICATION. 

MULTIPLICATION  is  a  compendious  method  of  performing  the 
operation  of  Addition.  It  consists  of  three  parts,  the  Multipli- 
cand, or  number  to  be  multiplied  ;  the  Multiplier,  or  number  by 
which  to  multiply ;  and  the  result,  which  is  called  the  Product. 
The  Multiplicand  and  Multiplier  are  called  factors. 

RULE. 

Place  the  larger  number  uppermost  for  the  multiplicand,  and  the 
smaller  number  under  it  for  a  multiplier,  arranging  units  under  units, 
tens  under  tens,  <5fc.  Then  multiply  each  figure  of  the  multiplicand  by 
each  figure  of  the  multiplier,  beginning  with  the  right-hand  figure,  and 
carrying  for  every  ten  as  in  addition.  Jf  the  multiplier  consists  of  more 
than  one  figure,  the  right-hand  figure  of  each  product  must  be  placed  di- 
rectly under  the  figure  of  the  multiplier  that  produces  it,  which  will  cause 
the  successive  products  to  recede  each  one  place  to  the  left.  The  sum  of 
the  several  products  will  be  the  whole  product  required. 

NOTE  ] .  —  When  there  are  ciphers  between  the  significant  figures  of 
the  multiplier,  pass  over  them  in  the  operation,  and  multiply  by  the 


SECT,  ir.j  MULTIPLICATION.  25 

significant  figures  only,  remembering  to  set  the  first  figure  of  the  product 
directly  under  the  figure  of  the  multiplier  that  produces  it.     See  Lx.  15. 

NOTE.  2.  —  If  there  are  ciphers  at  the  right  hand  either  of  the  multi- 
plier or  multiplicand,  or  of  both,  they  may  be  neglected  to  the.  close  of  the 
operation,  when  they  must  be  annexed  to  the  product. 

PROOF. 

The  correctness  of  the  result  in  Multiplication  may  be 
conveniently  ascertained  in  three  ways  ;  viz.,  by  Division,  by 
Multiplication,  or  by  casting  out  the  nines. 

According  to  the  first  method,*  divide  the  product  by  the  mul- 
tiplier ;  and,  if  the  work  is  right,  the  quotient  will  be  equal  to 
the  multiplicand. 

According  to  the  second  method,  take  the  multiplier  for  the 
multiplicand  and  the  multiplicand  for  the  multiplier,  and  pro- 
ceed according  to  the  rule  for  multiplication  ;  and,  if  the  work 
be  right,  the  product  will  be  the  same  as  by  the  former  opera- 
tion. 

According  to  the  third  method,  begin  at  the  left  hand  of  the 
multiplicand,  and  add  together  its  successive  figures  towards 
the  right,  till  the  sum  obtained  equals  or  exceeds  the  number 
9.  If  it  equals  it,  drop  the  nine,  and  begin  to  add  again  at  this 
point,  and  proceed  till  you  obtain  a  sum  equal  to  or  greater 
than  nine.  If  it  exceeds  nine,  drop  the  nine  as  before,  and  car- 
ry the  excess  to  the  next  figure,  and  then  continue  the  addition 
as  before.  Proceed  in  this  way  till  you  have  added  all  the 
figures  in  the  multiplicand  and  rejected  all  the  nines  contained 
in  it,  and  write  the  final  excess  at  the  right  hand  of  the  multipli- 
cand. Proceed  in  the  same  manner  with  the  multiplier,  and 
write  the  final  excess  under  that  of  the  multiplicand.  Multiply 
these  excesses  together  and  place  the  excess  of  nines  in  their 
product  under  the  other  excesses.  Then  proceed  to  find  the 
excess  of  nines  in  the  product  obtained  by  the  original  operation, 
and,  if  the  work  be  right,  the  excess  thus  found  will  be  equal  to 
the  excess  contained  in  the  product  of  the  above  excesses  of 
the  multiplicand  and  multiplier.  See  Example  15. 

NOTE.  —  This  method  of  proof,  though  perhaps  sufficiently  sure  for 
common  purposes,  is  not  always  a  test  of  the  correctness  of  an  operation. 
Cases  will  sometimes  occur  in  which  the  excesses  above  named  will  be 
equal,  when  the  work  is  not  right. 

*  As  the  pupil  is  presumed  not  to  be  acquainted  with  Division,  he  will 
pass  over  this  method  of  proof  for  the  present.  It  is  placed  here  as  a 
method  important  to  be  known,  and  because  there  seems  to  be  no  better 
place  for  it,  though  it  presupposes  an  acquaintance  with  a  rule  yet  to  be 
learned. 

3 


26 


MULTIPLICATION. 


[SECT.  iv. 


TABLE  OF  PYTHAGORAS. 


toto~. 


33 


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5 


fe  £5 


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s  t 


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'-  _* 


22fe 

,"ir*    rtsl  ^» 


38 


s:  ri  is 


sis 


re  |O  •"• 
*+  tv  co 

tCiOlGC 


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s  g 


I.I 


i: 


5SIS 

afs 


EXAMPLES. 
2.  3. 

36785678    123456789 
7  6 


678956324     36785678    123456789     987654321 
3    7     6    9 

2036868972    257499746     740740734    8888888889 


SECT,  ir.] 


MULTIPLICATION. 


27 


2715816 
1357908 

16294896 


3700698 
1850349 

22204188 


5526941 
3947815 

45005091 


12. 

789567 

98 

6316536 
7106103 

77377566 


13.  14.  15. 

3678543  67854  612346  =  4 

4567  10234  430049  =  2 

25749801  271416  5511114  =  8 

22071258  203562  2449384 

18392715  135708  1837038 

14714172  67854       2449384 

16799905881  694417836  263338784954=8 


16. 

678567 
8007 


17. 

4567895 
60004 


18. 

6785000 
32000 


19. 

478763000 
12000 


20.  Multiply 

21.  Multiply 

22.  Multiply 

23.  Multiply 

24.  Multiply 

25.  Multiply 

26.  Multiply 

27.  Multiply 

28.  Multiply 

29.  Multiply 

30.  Multiply 

31.  Multiply 

32.  Multiply 

33.  Multiply 

34.  Multiply 

35.  Multiply 


75432  by  47. 
76785316  by  7615. 
67853000  by  8765. 
38123450  by  31243. 
40670007  by  10002. 
31235678  by  10203. 
76786321  by  30070. 
317160070  by  700500. 
467325812  by  167000. 
6176777  by  22222. 


Ans.  3545304. 

Ans.  584720181340. 

Ans.  594731545000. 

Ans.  1191090948350. 

Ans.  406781410014. 

Ans.  318697622634. 

Ans.  2308964672470. 

Ans.  222170629035000. 

Ans.  78043410604000. 

Ans.  137260338494. 


123456789  by  987654321. 

Ans.  121932631112635269. 

7060504  by  30204.  Ans.  213255462816. 

6017853  by  800070.  Ans.  4814703649710. 

5000700  by  530071.  Ans.  2650726049700. 

88888  by  4444.  Ans.  395018272. 

123000  by  78000.  Ans.  9594000000. 


28  MULTIPLICATION.  [UKCT.  ir. 

36.  Multiply  7008005  by  10008.  Ans.  70136114040. 

37.  Multiply  4001100  by  40506.  Ans.   162068556600. 

38.  Multiply  6716700  by  808070.  Ans.  5427563769000. 

39.  Multiply  987648  by  481007.  Ans.  475065601536. 

40.  Multiply  187 11000  by  470.  Ans.  8794170000. 

41.  Multiply  10000  by  7000.  Ans.  70000000. 

42.  Multiply  101010101  by  2020202.  Ans.  204060808060402. 

43.  Multiply  70QOO  by  10000.  Ans.  700000000. 

44.  Multiply  800008  by  9009.  Ans.  7207272072. 

45.  Multiply  900900  by  70070.  Ans.  63126063000. 

46.  Multiply  4807658  by  706007.  Ans.  3394240201606. 

47.  Multiply  16789001  by  10080.  Ans.   169233130080. 

48.  Multiply  304050607  by  3011101.  Ans.  915527086788307. 

49.  Multiply  908007004  by  500123.  Ans.  454115186861492. 

50.  Multiply  2003007001  by  6007023. 

Ans.  12032109124168023. 

51.  Multiply  9000006  by  9000006.   Ans.  81000108000036. 

52.  Multiply  11 5292 1504606846976  by  1152921504606846976. 

Ans.  1329227995784915872903807060280344576. 

53.  What  will  27  oxen  cost  at  35  dollars  each  ? 

Ans.  $  945. 

54.  What  will  365  acres  of  land  cost  at  73  dollars  per  acre,? 

Ans.  $  26645. 

55.  What  will  97  tons  of  iron  cost  at  57  dollars  a  ton  ? 

Ans.  $  5529. 

56.  What  will  397  yards  of  cloth  cost  at  7  dollars  per  yard  ? 

Ans.  $2779. 

57.  What  will  569  hogsheads  of  molasses  cost  at  37  dollars 
per  hogshead  ?  Ans.  $21053. 

58.  If  a  man  travel  37  miles  in  one  day,  how  far  will  he 
travel  in  365  days  ?  Ans.   13505  miles. 

59.  If  one  quire  of  paper  have  24  sheets,  how  many  sheets 
are  in  a  ream,  which  consists  of  20  quires  ?  Ans.  480  sheets. 

60.  If  a  vessel  sails  169  miles  in  one  day,  how  far  will  she 
sail  in  144  days  ?  Ans.  24336  miles. 

61.  What  will  698  barrels  of  flour  cost  at  7  dollars  a  barrel? 

Ans.  $  4886. 

62.  What  will  376  Ibs.  of  sugar  cost  at  13  cents  a  pound  ? 

Ans.  4888  cts. 

63.  What  will  97  Ibs.  of  tea  cost  at  93  cents  a  pound  ? 

Ans.  9021  cts. 

64.  If  a  regiment  of  soldiers  consists  of  1 128  men,  how 
many  men  are  there  in  an  army  of  53  regiments  ? 

Ans.  59784. 


SECT,  v.]  DIVISION.  29 

65.  What  will  an  ox  weighing  569  pounds  amount  to  at  8 
cents  a  pound  ?  Ans.  4552  cts. 

66.  If  a  barrel  of  cider  can  be  bought  for  93  cents,  what  will 
75  barrels  cost  ?  Ans.  6975  cts. 

67.  If  in  a  certain  factory  786  yards  of  cloth  are  made  in  one 
day,  how  many  will  be  made  in  313  days  ?   Ans.  246018  yds. 

68.  A  certain  house  contains  87  windows,  and  each  window 
has  32  squares  of  glass ;  how  many  squares  are  there  in  the 
whole  house  ?  Ans.  2784  squares. 

69.  There  are  407  wagons  each  loaded  with  30009  pounds 
of  coal ;  how  many  pounds  are  there  in  the  whole  ? 

Ans.  122 13663  pounds. 

70.  Multiply  three   hundred  and   seventy-five  millions  two 
hundred  and  ninety-six  thousand  three  hundred  and  twenty-one, 
by  seventy-nine  thousand  and  twenty-four. 

Ans.  29657416470704. 

71.  What  would  be  the  cost  of  687  fothers  of  lead  at  73  dol- 
lars a  fother  ?  Ans.  $  50151. 


SECTION  V. 
DIVISION. 

THE  object  of  Division  is  to  find  how  many  times  one  num- 
ber is  contained  in  another. 

Division  consists  of  three  principal  parts ;  the  Dividend,  or 
number  to  be  divided ;  the  Divisor,  or  number  by  which  we  di- 
vide ;  and  the  Quotient,  which  shows  how  many  times  the  divi- 
dend contains  the  divisor. 

When  the  dividend  contains  the  divisor  an  exact  number  of 
times,  the  quotient  is  expressed  by  a  whole  number.  But  when 
this  is  not  the  case,  'there  will  be  a  remainder,  when  the  divis- 
ion has  reached  its  limit,  and  this  remainder  placed  above  the 
divisor,  with  a  horizontal  line  between  them,  will  form  a  frac- 
tion, and  should  be  written  at  the  right  hand  of  the  quotient, 
and  will  be  a  part  of  it.  See  Example  2d,  and  note. 

1.  The  Remainder  may  be  considered  a  fourth  term  in  Di- 
vision, and  it  will  always  be  of  the  same  denomination  with  the 
dividend. 

For  the  sake  of  convenience,  Division  has  been  divided  into 
two  kinds,  Long  and  Short. 
3* 


30  DIVISION.  [SECT.  T. 

2.  All  questions  in  which  the  divisor  is  not  more  than  12 
may  be  conveniently  performed  by  Short  Division  ;  all  others 
are  better  performed  by  Long  Division. 

SHORT  DIVISION. 

EXAMPLE. 

1.  Divide  948  dollars  equally  among  4  men- 

n  • ,  v/    ,/         I*1  performing  this  question,  inquire  how 
D'  '       4^948         many  times  4,  the  divisor,  is  contained  in  9, 
which  is  2  times,  and  1   remaining ;  write 
Quotient    237         fae  2  under  the  9  and  suppose  i?  tne  re-. 

mainder,  to  be  placed  before  the  next  figure  of  the  dividend,  4, 
and  the  number  will  be  14.  Then  inquire  how  many  times 
4,  the  divisor,  is  contained  in  14.  It  is  found  to  be  3  times  and 
2  remaining.  Write  the  3  under  the  4,  and  suppose  the  re- 
mainder, 2,  to  be  placed  before  the  next  figure  of  the  dividend, 
8,  and  the  number  will  be  28.  Inquire  again  how  many 
times  28  will  contain  the  divisor.  It  is  found  to  be  7  times, 
which  we  place  under  the  8.  Thus  we  find  each  man  receives 
237  dollars. 

From  the  above  illustration,  we  deduce  the  following 

RULE. 

Write  down  the  dividend  and  place  the  divisor  on  the  left,  with  a 
curved  or  perpendicular  line  drawn  between  them.  Draw  also  a  hori- 
zontal line  under  the  dividend,  then  observe  hoiv  many  times  the  divisor 
is  contained  in  the  first  figure  or  figures  of  tlie  dividend  (beginning  at 
the  left  hand) ,  and  place  the  quotient  figure  directly  under  the  right-hand 
figure  of  the  part  of  tlie  dividend  that  was  taken.  If  there  be  no  re- 
mainder, proceed  to  inquire  how  many  times  the  divisor  is  contained  in 
the  next  figure  *  of  the  dividend,  and  set  down  the  result  at  the  right 
hand  of  the  quotient  figure  already  obtained,  or  directly  under  the  figure 
of  the  dividend,  and  continue  the  work  in  this^nanner  until  the  whole 
dividend  is  divided.  But  if  there  be  a  remainder  either  in  the  first  or 
any  subsequent  division,  imagine  the  number  denoting  it  to  be  placed  di- 
rectly before  the  next  figure  of  the  dividend,  and  ascertain  the  number  of 
times  the  divisor  is  contained  in  the  number  thus  formed,  and  place  the 

*  If  this  figure  be  smaller  than  the  divisor,  it  cannot  contain  it,  and  the 
figure  to  be  placed  in  the  quotient  will  be  a  cipher.  Sometimes,  as  when 
we  divide  by  11  or  12,  we  may  have  two  successive  ciphers  in  the  quo- 
tient, as  when  the  divisor  is  12  and  the  next  two  figures  are  1  's  or  1  and  0. 
We  are  then  obliged  to  proceed  to  a  third  figure  in  the  dividend,  before 
we  can  effect  a  proper  division. 


SECT.  V.J 


DIVISION. 


31 


quotient  figure  underneath,  as  before.  Proceed  in  this  way  until  every 
part  of  the  dividend  is  thus  divided,  and  the  result  will  be  the  quotient 
sought. 


3)67856336 

22618778§* 


EXAMPLES. 
3. 

5)123456789 
24691357$ 


6)98786356 


7)126711001 


8)33445567 


9)1234567 


10)178985 


11)1667789 


10. 
12)9167856 


Quotients. 

Rem. 

11.  Divide 

67893536  by  2 

33946768 

12.  Divide 

316789311  by  3 

105596437 

13.  Divide 

567895326  by  4 

141973831 

2 

14.  Divide 

123456789  by  5 

24691357 

4 

15.  Divide 

671678953  by  6 

111946492 

1 

16.  Divide 

166336711  by  7 

23762387 

2 

17.  Divide 

161331793  by  8 

20166474 

1 

18.  Divide 

161677678  by  9 

17964186 

4 

19.  Divide 

363895678  by  11 

33081425 

3 

20.  Divide 

164378956  by  12 

13698246 

4 

21.  Divide 

78950077  by  3 

1 

22.  Divide 

678956671  by  4 

3 

23.  Divide 

667788976  by  5 

1 

24.  Divide 

777777777  by  6 

3 

25.  Divide 

888888888  by  7 

6 

26.  Divide 

999999999  by  8 

7 

27.  Divide 

100000000  by  7 

2 

*  From  this  and  subsequent  examples  it  will  be  seen  that  fractions 
arise  from  division,  and  are  parts  of  a  unit ;  that  the  denominator  of  the 
fraction  represents  the  divisor,  and  shows  into  how  many  parts  the  given 
number  or  quantity  is  divided,  and  the  numerator,  being  the  remainder, 
shows  how  many  units  of  the  given  quantity  or  dividend  remain  undi- 
vided. By  writing  the  numerator  over  the  denominator  in  the  form  of  a 
fraction,  we  signify  that  it  is  to  be  divided  by  the  denominator  ',  and  when 
placed  at  the  right  hand  of  the  whole  number  in  the  quotient,  the  fraction 
becomes  a  part  of  the  quotient,  and,  as  such,  is  as  much  less  than  a  unit, 
as  the  numerator  is  less  than  the  denominator. 


32  DIVISION.  [SECT.  r. 

LONG  DIVISION. 

CASE  I. 

EXAMPLE. 

1.  A  prize,  valued  at  $3978,  is  to  be  equally  divided  among 
17  men.     What  is  the  share  of  each  ? 

The  object  of  this  ques- 

OPERATION.  tjon  jg    to     find  h()W  many 

Dividend.  times   3973  will  contain 

Divisor.  17)  3978  (  234  Quotient.  17?  or  how  many  times 

must    17    be     subtracted 

57     1638  from  3978,  until  nothing 

51     234  shall   remain.      We  first 

1J8  3978  Proof.  inquire,  how  many  times 

68  the   first  two  figures   of 

ID        •  **«.          tne  dividend  will  contain 
00  Reminder.  .  how 


many  times  39  will  contain  17.  Having  found  it  to  be  2  times, 
we  write  2  in  the  quotient  and  multiply  the  divisor,  17,  by  it, 
and  place  their  product  34  under  39,  from  which  we  sub- 
tract it,  and  find  the  remainder  to  be  5,  to  which  we  annex  the 
next  figure  of  the  dividend,  7.  And  having  found  that  57  will 
contain  the  divisor  3  times,  we  write  3  in  the  quotient,  multiply 
it  by  17,  and  place  the  product  51  under  57,  from  which  we 
subtract  it,  and  to  the  remainder,  6,  we  annex  the  next  figure 
of  the  dividend,  8,  and  inquire  how  many  times  68  will  contain 
the  divisor,  and  find  it  to  be  4  times.  And  having  placed  the 
product  of  4  times  17  under  68,  we  find  there  is  no  remainder, 
and  that  3978  will  contain  17,  the  divisor,  234  times  ;  that  is, 
each  man  will  receive  234  dollars.  To  prove  our  work  is  right, 
we  reason  thus.  If  one  man  receives  234  dollars,  17  men  will 
receive  17  times  as  much,  and  17  times  234  are  3978,  the  same 
as  the  dividend  ;  and  this  operation  is  effected  by  multiplying 
the  divisor  by  the  quotient.  The  student  will  now  see  the  pro- 
priety of  the  following 

RULE 

Place  tJie  divisor  and  dividend  as  under  the  preceding  rule,  and  draw 
a  curved  or  perpendicular  line  on  the  right  of  the  dividend.  Then  ob- 
serve how  many  figures  of  the  dividend,  counting  from  left  to  right, 
must  be  taken  to  contain  the  divisor  one  or  more  times,  but  never  ex- 
ceeding nine  times,  and  ascertain  how  many  times  these  figures  will 


contain  the  divisor,  placing  the  quotient  figure  on 
dividend.  Then  multiply  the  divisor  by  this  quotient 
the  product  in  order  under  t/ie  figures  of  the  dividend 
Subtract  this  product  from  the  part  of  the  dividend  above  it,  and  to  the 
difference  bring  down  and  annex  the  next  figure  of  the  dividend.  Di- 
vide this  number  by  the  divisor ,  and  place  the  quotient  figure  on  the  right 
of  the  one  already  found.  Multiply  the  divisor  by  the  quotient  figure 
last  found,  and  subtract  the  product  from  the  number  last  divided,  and 
bring  down  and  annex  as  before,  till  the  last  figure  of  the  dividend  is 
taken  ;  and  the  several  figures  on  tfte  right  of  the  dividend  will  be  the 
quotient  required.  The  difference  between  the  number  last  divided  and 
the  last  product  will  be  the  remainder,  which,  with  the  divisor,  will  form 
a  fraction,  as  under  the  preceding  rule. 

NOTE  1.  —  It  will  often  happen,  that,  after  a  figure  is  brought  down 
and  annexed  to  a  remainder,  the  number  will  not  contain  a  divisor. 
In  such  a  case,  a  cipher  is  to  be  placed  in  the  quotient,  and  the  next  figure 
to  be  brought  down  and  annexed,  and  thus  till  the  number  formed  shall 
be  large  enough  to  contain  the  divisor.  Sometimes  it  will  be  necessary 
thus  to  place  several  ciphers  in  succession  in  the  quotient. 

NOTE  2.  —  The  proper  remainder  is  in  all  cases  less  than  the  divisor; 
and  if,  at  any  time,  the  subtraction  named  in  the  rule  gives  a  remainder 
larger  than  the  divisor,  we  discover  at  once,  that  an  error  has  been  com- 
mitted in  the  division,  and  that  the  quotient  figure  must  be  increased. 

PROOF. 

Division  may  be  proved  by  Multiplication,  by  Addition,  by 
casting  out  the  9's,  or  by  Division. 

By  the  first  method,  we  multiply  the  quotient  by  the  divisor, 
adding  to  the  product  the  remainder,  and  the  result,  if  the  work 
be  right,  is  equal  to  the  dividend. 

By  the  second  method,  we  add  up  the  several  products  of  the 
several  quotient  figures  by  the  divisor,  together  with  the  re- 
mainder, and  the  result,  if  the  work  is  right,  is  like  the  divi- 
dend. See  Example  2. 

To  prove  Division  by  casting  out  the  9's,  we  find  the  excess 
of  9's  in  the  divisor  and  also  in  the  quotient,  and  multiply  these 
excesses  together  and  find  the  excess  in  their  product.  We 
then  subtract  the  remainder  from  the  dividend,  and  find  the  ex- 
cess of  9's  in  the  difference,  which,  if  the  work  is  right,  will 
be  equal  to  the  excess  found  in  the  product  of  the  excesses 
above  named.  See  Example  3. 

To  prove  Division  by  Division  itself,  we  subtract  the  remain- 
der from  the  dividend,  and  divide  the  difference  by  the  quotient, 
and,  if  the  work  is  right,  the  result  will  be  equal  to  the  original 
divisor.  See  Example  4. 


34 


DIVISION. 


[SECT.  r. 


EXAMPLES. 


97)147856(1524 
97 
508 
485* 
235 
194* 

416 

388* 


328)678767(2069 
656 


Proof. 

5 
4x8 

5 


28* 

147856 

*  NOTE.  —The  asterisks  show  the 
numbers  to  be  added. 


72)37895(526 
360 

Proof. 

37895 
23 

189 
144 
455 
432 

526)37872(72 
3682 
1052 
1052 

23 

Quotients. 

Bern. 

5.  Divide 

6756785  by 

35 

193051 

6.  Divide 

789636  by 

46 

17166 

7.  Divide 

7967848  by 

52 

153227 

44 

8.  Divide 

16785675  by 

61 

275175 

9.  Divide 

675753  by 

39 

17327 

10.  Divide 

5678910  by 

82 

69255 

11.  Divide 

6716394  by 

94 

71451 

12.  Divide 

1167861  by 

135 

8650 

111 

13.  Divide 

7861783  by 

87 

90365 

28 

14.  Divide 

1678567  by 

365 

4598 

297 

15.  Divide 

87635163  by 

387 

226447 

174 

16.  Divide 

34567890  by 

6789 

5091 

5091 

17.  Divide 

78911007  by 

36712 

2149 

16919 

18.  Divide 

78963167  by 

45671 

1728 

43679 

19.  Divide 

671616589  by 

61476 

10924 

52765 

20.  Divide 

471361876  by 

36789 

12812 

21208 

21.  Divide 

300700801  "by 

10037 

29959 

2318 

22.  Divide 

10000000  by 

9999 

1000 

1000 

23.  Divide 

199999999  by 

123456 

1279 

SECT.  T.] 


DIVISION. 


35 


Rem. 

24.  Divide  6716789513  by  7816789     2167762 

25.  Divide  1613716131  by  3151638  77475 

26.  Divide  121932631112635269  by  123456789 

27.  Divide  213255467083  by  30204  4267 

28.  Divide  4814703652065  by  800070  2355 

29.  Divide  2650726050934  by  530071  1234 

30.  Divide  395020613  b>  4444  2341 

31.  Divide  9594004321  by  78000  4321 

32.  Divide  162068563389  by  40506  6789 

33.  Divide  5427563776896  by  808070  7896 

34.  Divide  475065610503  by  481007  8967 

35.  Divide  8794170278  by  470  278 

36.  Divide  70006876  by  7000  6876 

37.  Divide  204060808062747  by  2020202  2345 

38.  Divide  700003456  by  10000  3456 

39.  Divide  7207276639  by  9009  4567 

40.  Divide  63126068678  by  70070  5678 

41.  Divide  3394240208391  by  706007  6785 

42.  Divide  169233137936  by  10080  7856 

43.  Divide  915527086796874  by  3011101  8567 

44.  Divide  454115186870257  by  500123  8765 

45.  Divide  12032109124169380  by  6007023  1357 

CASE  II. 

To  divide  by  any  number  with  ciphers  annexed. 
Cut  off  the  ciphers  from  the  divisor,  and  the  same  number  of  figures 
from  the  right  hand  of  the  dividend.  Then  divide  the  remaining  fig- 
ures of  the  dividend  by  the  remaining  figures  of  the  divisor,  and  the  re- 
sult will  be  the  quotient.  To  complete  the  work,  annex  to  the  last  re- 
mainder found  by  the  operation  the  figures  cut  off  from  the  dividend , 
and  the  whole  will  form  the  true  remainder. 

EXAMPLE. 

1.  Divide  36378967  by  31000. 

31,000)36378,967(1173 
31 

~53 
31 

227 
217 
108 
93 
15967  Remainder. 


36  DIVISION.  [SKCT.  y. 

Quotients.  Rem. 

2  Divide  32100  by    6000  5  2100 

3.  Divide  3167810  by  160000  19  127810 

4.  Divide  12345678  by  1400000  8  1145678 

5.  Divide  1637851  by  500000  3  137851 

6.  Divide  3678953  by  326100  11  91853 

7.  Divide  41111111  by  1100000  37  411111 

CASE  III. 

To  divide  by  a  unit  with  ciphers  annexed. 

Cut  off  as  many  figures  from  the  right  hand  of  the  dividend  as  there 
are  ciphers  in  the  divisor,  and  the  figures  on  the  left  hand  of  the  separa- 
trix  will  be  the  quotient,  and  those  on  the  right  hand  the  remainder. 

Quotients.  Rem. 

1.  Divide  123456789  by  10  12345678     9 

2.  Divide  987654321  by  100  9876543    21 

3.  Divide  1221 12347800  by  1000  122112347    800 

4.  Divide  89765432156  by  1000000  89765  432156 

CASE  IV. 

To  divide  by  a  composite  number,  that  is,  a  number  pro- 
duced by  the  multiplication  of  two  or  more  numbers. 

Divide  the  dividend  by  any  one  of  the  factors,  and  the  quotient  thus 
found  by  anotJier,  and  thus  proceed  till  every  factor  has  been  made  a  di- 
visor, and  the  last  quotient  will  be  the  true  quotient  required. 

NOTE.  —  To  find  the  true  remainder,  we  multiply  the  last  remainder 
by  the  last  divisor  but  one,  and  to  the  product  add  the  next  preceding 
remainder ;  we  multiply  this  sum  by  the  next  preceding  divisor,  and  to 
the  product  add  the  next  preceding  remainder;  and  so  on,  till  we  have 
gone  through  all  the  divisors  and  remainders  to  the  first. 

This  rule  will  be  better  understood  by  the  pupil,  after  he  has 
become  acquainted  with  fractions. 

EXAMPLES. 

1.  Divide  47932  by  72. 

As  72  is  equal  to  9  times  8,  we  first 
9)47932  divide  the  dividend  by  9,  and  the  quotient 

8^5325 7  thence  arising  by  8  ;  and  to  find  the  true 

' remainder,  we  multiply  the  last  remainder, 

665  5  r=  52  5^  by  the  first  divisor,  9,  and  to  the  prod- 
uct add  the  first  remainder,  7 ;  and  find 
the  amount  to  be  52,  the  true  remainder. 


SECT,  vi.]    CONTRACTIONS  IN  MULTIPLICATION.  37 

2.  Divide  5371  by  192. 

We  find  192  equal  to  the  product  of  4 

4)5371  times  6  times  8,  =  4x6x8=  192.  We 

6^1342  —  3  therefore  divide  by  these  factors,  as  in  the 

last  example.     To  find  the  true  remainder, 

8)223      4  we  multiply  the  last  remainder,  7,  by  the 

27  —  7  =  187  last  divisor  but  one,  6 ;  and  to  the  product 

add  the  last  remainder  but  one,  4 ;  this  sum 

we  multiply  by  the  first  divisor,  4 ;  and  to  the  product  add  the 

first  remainder,  3  ;  and  find  the  amount  to  be  187. 

Quotients.       Rem. 

3.  Divide  7691  by    24=   4  X    6  320  11 

4.  Divide  8317  by    27  =   3  X    9  308  1 

5.  Divide  3116  by    81  =   9  X    9  38  38 

6.  Divide  61387  by  121  =  11  x  11  507  40 

7.  Divide  19917  by  144  =  12  X  12  138  45 

8.  Divide  9 1746  by  336=    6x    7X8  273  18 

9.  Divide  376785  by  315  =   5  X    7x9  1196  45 


SECTION  VI. 

CONTRACTIONS  IN  MULTIPLICATION. 

I.  To  multiply  by  25. 

RULE.  —  Annex  two  ciphers  to  the  multiplicand,  and  divide  it  by  4, 
and  the  quotient  is  the  product  required. 

Rationale.  —  By  annexing  two  ciphers,  we  increase  the  mul- 
tiplicand one  hundred  times,  and  by  dividing  this  number  by  4, 
the  result  will  be  an  increase  of  the  multiplicand  only  twenty- 
five  times,  because  25  is  one  fourth  of  100. 

1.  Multiply  785643  by  25. 

OPERATION. 

4)78564300 

19641075  Product. 

2.  Multiply  9876543  by  25.  Ans. 

3.  Multiply  47110721  by  25.  Ans. 

II.  To  multiply  by  33  J. 

RULE.  —  Annex  two  ciphers  to  the  multiplicand,  and  divide  it  by  3, 
and  the  quotient  is  the  product  required. 
4 


38  CONTRACTIONS  IN  MULTIPLICATION.      [SECT.  vi. 

Rationale.  —  As  in  the  last  case,  by  annexing  two  ciphers,  we 
increase  the  multiplicand  one  hundred  times ;  and  by  dividing 
the  number  by  3,  we  only  increase  the  multiplicand  thirty-three 
and  one  third  times,  because  33£  is  one  third  of  100. 

4.  Multiply  87138942  by  33£. 

OPERATION. 

3)8713894200 
2904631400  Product. 

5.  Multiply  66666993  by  33£.  Ans. 

6.  Multiply  12336723  by  33£.  Ans. 

EL  To  multiply  by  125. 

RULE. — Annex  three  ciphers  to  the  multiplicand,  and  divide  by  8, 
and  the  quotient  is  the  product. 

NOTE.  —  By  annexing  three  ciphers,  the  number  is  increased  one  thou- 
sand times;  and,  by  dividing  by  8,  the  quotient  will  be  only  one  eighth 
.  of  1000,  that  is,  125  times. 

7.  Multiply  12345678  by  125. 

OPERATION. 

8)12345678000 

1543209750  Product. 

IV.  To  multiply  by  any  number  of  9's. 

RULE.  —  Annex  as  many  ciphers  to  the  multiplicand  as  there  are  9's 
in  the  multiplier,  and  from  this  number  subtract  the  number  to  be  multi- 
plied, and  the  remainder  is  the  product  required. 

8.  Multiply  87654  by  999. 

OPERATION.  By  annexing  three  ciphers,  we  make  the 

87654000  number  one  thousand  times  larger.     If  from 

87654  ^s  nu^ber,  with  the  ciphers  annexed,  we 

p^j     ,    subtract  the  multiplicand,  we  make  the  prod- 
Product.  uct  one  thousandth  part   less;  that  £  the 

product  will  be  only  999  times  the  multiplicand.     Q.  E.  D. 

9.  Multiply  7777777  by  9999.  Ans.  77769992223. 

10.  Multiply  5555  by  999999.  Ans.  555499-1445. 

•I*0,!*"""1?0  m"ltiP1y  b7  any  number  of  3's,  proceed  as  above  and  di- 
vide the  product  by  3;  but  if  it  be  required  to  multiply  by  6's,  proceed  as 
above  and  then  multiply  the  product  by  2,  and  divide  the" result  by  3,  and 
the  quotient  is  the  product. 


SECT,  vii.]  CONTRACTIONS  IN  DIVISION. 

11.  Multiply  987654  by  333333. 

OPERATION. 

987654000000 
987654 


3)987653012346 

329217670782  Product,  Ans. 

12.  Multiply  32567895  by  3333.  Ans. 

13.  Multiply  876543  by  66666.  Ans. 

OPERATION. 

87654300000 

876543 

87653423457 

2 

3)175306846914 

58435615638  Product,  Ans. 

14.  Multiply  345678  by  6666666.  Ans. 

V.  When  the  multiplier  can  be  separated  into  periods,  which 
are  multiples  of  one  another,  the  operation  may  be  contracted 
in  the  following  manner. 

15.  Multiply  112345678  by  288144486. 

OPERATION. 

112345678 
288144486 

674074068  =  the  product  by  6. 
5392592544    =  the  foregoing  product  X  by  8  for  48. 
16177777632        =  the  lest  product  x  by  3  for  144. 

32355555264 =  the  last  product  X  by  2  for  288. 

32371787641631508  Product. 


SECTION  VII. 

CONTRACTIONS  IN  DIVISION. 

I.  To  divide  by  5. 

RULE.  —  Multiply  the  dividend  by  2,  and  tJie  product,  except  the  last 
figure  at  the  right,  is  the  quotient. 

NOTE.  —  The  remainder  will  be  tenths. 


40  CONTRACTIONS  IN  DIVISION.  [SECT.  VH. 

1.  Divide  67895  by  5.  Ans.  13579. 

OPERATION. 

67895 

2 

13579,0  Quotient. 

II.  To  divide  by  25. 

RULE.  —  Multiply  the  dividend  by  4,  and  the  product,  except  the  last 
two  figures  at  the  right,  is  the  quotient. 

NOTE.  —  The  two  figures  at  the  right  are  hundredths. 

2.  Divide  8765887  by  25.  Ans.  350635^. 

OPERATION. 

8765887 

4 

350635,48  Quotient. 

HI.  To  divide  by  33  J. 

RULE.  —  Multiply  the  dividend  by  3,  and  the  product,  except  the  last 
two  figures  at  the  right,  is  the  quotient,  and  the  last  two  are  hundredths. 

3.  Divide  876735  by  33£.  Ans.  26302^. 

OPERATION. 

876735 
3 

26302,05  Quotient. 

IV.  To  divide  by  125. 

RULE.  —Multiply  the  dividend  by  8,  and  the  product,  except  the  last 
three  figures,  is  the  quotient,  and  these  last  three  figures  will  be  thou- 
sandths. 

4.  Divide  1234567  by  125.  Ans. 

OPERATION. 

1234567 

8 


9876,536  Quotient. 

5.  Divide  8786789  by  125.  Ans 

6.  Divide  1234567  by  125  Ans. 

V.  A  short  method  of  performing  Long  Division. 

7.  Divide  16294996  by  24.  Ans.  6789M. 


SECT,  vin.]  MISCELLANEOUS  EXAMPLES.  41 

OPERATION.  This  method   differs   from   the 

24)16294896(678954,  Ans.  common  way  byplacing  the  right- 

hand  figure  of  every  product  im- 
181229  mediately  under  the  dividend. 

169129 
121 
121 

8.  Divide  3545304  by  47.  Ans.  75432. 

9.  Divide  45005091  by  57.  Ans.  789563. 

VI.  To  divide  by  any  number  of  9's,  when  their  number  is 
not  less  than  half  the  number  of  places  that  will  be  in  the  quo- 
tient, and  when  there  is  no  remainder. 

RULE.  — Annex  as  many  ciphers  to  the  dividend,  as  there  are  9's  in  the 
divisor.  Then  write  the  proper  dividend  under  the  number  thus  found, 
and  subtract  it  from  the  number  to  which  ciphers  have  been  annexed; 
and,  as  many  places  of  the  remainder  at  the  right  hand  as  there  were 
ciphers  annexed,  are  so  many  figures  for  the  right  hand  of  the  quotient ; 
and,  for  t/ie  remaining  numbers  of  the  quotient,  a  competent  number 
must  be  taken  from  the  left  hand  of  the  above  remainder. 

10.  Divide  123332544  by  999. 

OPERATION.  By-   examining  the  dividend  and 

123332544000  divisor,  we  know   there  will   be  6 

123332544  places  in  the  quotient.     We  there- 

123  209211  456  ta^e  t^lree  °^  mese  figures  from 

I90d*fi  Our       t  An«  the  ri§ht  hand  of  the  remainder  for 
>  Quotient,  Ans.  ^  ^  right.hand  figureg  of  the 

quotient,  and  the  other  three  we  take  from  the  left  hand  of  the 
remainder. 

11.  Divide  12332655  by  999.  Ans.  12345. 

12.  Divide  987551235  by  9999.  Ans.  98765. 

13.  Divide  9123456779876543211  by  999999999. 

Ans.  9123456789. 


SECTION  VIII. 
MISCELLANEOUS  EXAMPLES. 

1.  What  number  multiplied  by  1728  will  produce  1705536  ? 

Ans.  987. 
4* 


42  MISCELLANEOUS  EXAMPLES.  [SECT.  vm. 

2.  If  a  garrison  of  987  men  are  supplied  with  175686  pounds 
of  beef,  how  much  will  there  be  for  each  man  ?    Ans.  178  Ibs. 

3.  In  one  dollar  there  are  100  cents ;  how  many  dollars  in 
697800  cents  ?  Ans.  $  6978. 

4.  In  one  pound  there  are  16  ounces ;  how  many  pounds  are 
in  111680  ounces  ?  Ans.  6980  Ibs. 

5.  A  dollar  contains  6  shillings ;  how  many  dollars  are  in 
5868  shillings  ?  Ans.  $  978. 

6.  The  President  of  the  United  States  receives  a  salary  of 
$  25,000 ;  what  does  he  receive  per  month  ?     Ans.  $  2083£. 

7.  A  man  receiving  $  96  for  8  months'  labor,  what  does  he 
receive  for  1  month  ?  Ans.  $  12. 

8.  The  distance  from  Haverhill  to  Boston  is  30  miles  ;  and,  if 
a  man  travel  6  miles  an  hour,  how  long  will  he  be  in  going  this 
distance  ?  Ans.  5  hours. 

9.  The  annual  revenue  of  a  gentleman  being  $8395,  how 
much  per  day  is  that  equivalent  to,  there  being  365  days  in  a 
year?  Ans.  $23. 

10.  The  car  on  the  Liverpool  railroad  goes  at  the  rate  of  65 
miles  an  hour ;  how  long  would  it  take  to  pass  round  the  globe, 
the  distance  being  about  25,000  miles  ?      Ans.  384^  hours. 

11.  How  much  sugar  at  $  15  per  cwt.  may  be  bought  for 
$  405  ?  Ans.  27  cwt. 

12.  In  6789560  shillings  how  many  pounds,  there  being  20 
shillings  in  a  pound  ?  Ans.  339478  pounds. 

13.  The  Bible  contains  31,173  verses  ;  how  many  must  be 
read  each  day,  that  the  book  may  be  read  through  in  a  year  ? 

Ans.  85-^|f  verses. 

14.  In  123456720  minutes  how  many  hours  ? 

Ans.  2057612  hours. 

15.  A  gentleman  possessing  an  estate  of  $  66,144,  bequeathed 
one  fourth  to  his  wife,  and  the  remainder  was  to  be  divided  be- 
tween his  4  children  ;  what  was  the  share  of  each  ? 

Ans.  $  12,402. 

16.  A  man  disposed  of  a  farm  containing  175  acres  at  $  87 
per  acre  ;  of  the  avails  he  distributed  $  1234  for  charitable  pur- 
poses ;  $  197  was  expended  for  the  purchase  of  a  horse  and 
chaise  ;  the  remainder  was  divided  between  6  gentlemen  and  8 
ladies,  and  each  lady  was  to  receive  twice  as  much  as  a  gentle- 
man ;  what  was  the  share  of  each  ? 

Ans.  $  627  for  a  gentleman,  and  $  1254  for  a  lady. 

17.  If  there  are  160  square  rods  in  an  acre,  how  many  acres 
are  in  1086240  square  rods  ?  Ans.  6789  acres. 


SECT,  vin.]  MISCELLANEOUS  EXAMPLES.  43 

18.  If  144  square  inches  make  one  square  foot,  how  many 
square  feet  in  14222160  square  inches  ?       Ans.  98765  feet. 

19.  What  number  is  that,  which  being  multiplied  by  24,  the 
product  divided  by  10,  the  quotient  multiplied  by  2,  32  sub- 
tracted from  the  product,  the  remainder  divided  by  4,  and  8 
subtracted  from  the  quotient,  the  remainder  shall  be  2  ? 

Ans.  15. 

20.  What  is  the  difference  between  half  a  dozen  dozen,  and 
six  dozen  dozen  ?  Ans.  792. 

21.  Bought  of  F.  Johnson  8  barrels  of  flour  at  $  7  per  barrel, 
and  3  hundred  weight  of  sugar  at  $  8  per  hundred.     What  was 
the  amount  of  his  bill  ?  Ans.  $  80. 

22.  Sold  S.  Jenkins  my  best  horse  for  $  75,  my  second-best 
chaise  for  $  87,  a  good  harness  for  $  31.     He  has  paid  me  in 
cash  $  38,  and  has  given  me  an  order  on  Peter  Parker  for  $  12. 
How  many  dollars  remain  my  due  ?  Ans.  $  143. 

23.  T.  Webster  has  sold  his  wagon  to  J.  Emerson  for  $  85. 
He  is  to  receive  his  pay  in  wood  at  $  5  per  cord.     How  many 
cords  will  it  require  to  balance  the  value  of  the  wagon  ? 

Ans.   17  cords. 

24.  Purchased  a  farm  of  500  acres  for  $  17,876.     I  sold  127 
acres  of  it  ut  $  47  an  acre,  212  acres  at  $  96  an  acre,  and  the 
remainder  at  $  37  an  acre.     What  did  I  gain  by  my  bargain  ? 

Ans.  $14,402. 

25.  A  tailor  has  938  yards  of  broadcloth  ;  how  many  cloaks 
can  be  made  of  the  cloth,  if  it  require  7  yards  to  make  one 
cloak  ?  Ans.   134  cloaks. 

26.  Bought  97  barrels  of  molasses  at  $  5  a  barrel.     Gave  17 
barrels  to  support  the  poor,  and  the  remainder  was  sold  at  $  8 
a  barrel.     Did  I  gain  or  lose,  and  how  much  ? 

Ans.  $  155  gain. 

27.  There  are  12  pence  in  one  shilling ;  required  the  num- 
ber of  pence  in  671  shillings.  Ans.  8052  pence. 

28.  Twelve   inches  make  one  foot  in  length ;  required  the 
number  of  inches  in  5280  feet,  it  being  the  length  of  a  mile. 

Ans.  63360  inches. 

29.  In  one  pound  avoirdupois  there  are  16  ounces ;   required 
the  ounces  in  1728  pounds.  Ans.  27648  ounces. 

30.  Required  the  number  of  shillings  in  8136  pence. 

Ans.  678  shillings. 

31.  It  requires  1728  cubic  inches  to  make  one  cubic  foot ; 
required  the  number  of  cubic  inches  in  3787  cubic  feet. 

Ans.  6543936  inches. 


44 


MONEY  AND  WEIGHTS. 


[SECT.  ix. 


SECTION  IX. 
TABLES  OF  MONEY,  WEIGHTS,  AND  MEASURES. 

T7NITED    STATES    MONEY. 

1  Cent,  marked  c. 

1  Dime,  "  d. 

1  Dollar,  "  $. 

1  Eagle,  "  E. 


10 
10 
10 
10 

Mills. 
10 
100 
1000 
10000 

Mills 
Cents 
Dimes 
Dollars 

n 

Cents. 
1 
10 
100 
1000 

Dimes. 

1  Dollars. 

10  =  1  Eagle. 

100  =  10         =       1 


4  Farthings 

12  Pence  " 

20  Shillings  " 

21  Shillings  sterling  " 
28  Shillings  N.  E.  " 


ENGLISH   MONEY. 

make 


1  Penny, 
1  Shilling, 
1  Pound, 
1  Guinea, 
1  Guinea, 


NOTE.  —  One  pound  sterling  is  equal  to  $  4.44|- 


4 

48 
960 


d. 
1 

12 
240 


marked 


a. 

1 

20 


d. 

s. 

£. 

G. 

G. 


100  Centimes 


FRENCH    MONEY. 

make       1  Franc        : 


.186  dollar. 


TROY   WEIGHT. 

24  Grains  make         1  Pennyweight,^    marked       dwt. 

20  Pennyweights  "  1  Ounce,  "  oz. 

12  Ounces  "  1  Pound,  Ib. 

gfr 

24 
480 
5760 
By  this  weight  are  weighed  gold,  silver,  and  jewels. 

NOTE. — "  The  original  of  all  weights  used  in  England  was  a  grain  or 
corn  of  wheat,  gathered  out  of  the  middle  ofthe  ear ;  and,  being  well  dried, 
32  of  them  were  to  make  one  pennyweight,  20  pennyweights  one  ounce, 
and  12  ounces  one  pound.  But  in  later  times,  it  was  thought  sufficient  to 
divide  the  same  pennyweight  into  24  equal  parts,  still  called  grains,  being 
the  least  weight  now  in  common  use  j  and  from  hence  the  rest  are  com- 
puted." 


make 

dwt. 
1 
20 
240 

1  Pennyweight,^ 
1  Ounce, 
1  Pound, 

oz. 

=          1 

=             12 

SECT,  ix.] 


WEIGHTS  AND  MEASURES. 


APOTHECARIES'  WEIGHT. 


20  Grains  S»\ 
3  Scruples 
8  Drams  S 

12  Ounces 


60 

480 
5760 


make 


1 

3 

24 

288 


1  Scruple, 
1  Dram, 
1  Ounce, 
1  Pound, 

dr. 

1 

8 

=  v       96 


marked  sc.  or  9 
"  dr.  or  5 
"  oz.  or 

"        Ib.  or 


oz. 

1 

12 


Apothecaries  mix  their  medicines  by  this  weight ;  but  buy  and  sell 
by  Avoirdupois.  The  pound  and  ounce  of  this  weight  are  the  same 
as  in  Troy  Weight. 

AVOIRDUPOIS   WEIGHT. 


16  Drams 

make 

1  Ounce, 

marked  oz. 

16  Ounces 

" 

1  Pound, 

Ib. 

28  Pounds 

«« 

, 

1  Quarter 

"        qr. 

4  Quarters 

« 

| 

1  Hundrec 

Weight, 

"     cwt. 

20  Hundred 

Weight 

H 

i 

1  Ton, 

"      ton. 

dr. 

oz. 

' 

16*     = 

1 

Ib. 

256       = 

16 

= 

1 

qr. 

7168      = 

448 

= 

| 

28      = 

cwt. 

28672       = 

1792 

= 

1 

112       = 

4        = 

1                 ton. 

573440       = 

35840 

=       2240       = 

80       = 

20       =       1 

I3y  this  weight  are  weighed  almost  every  kind  of  goods,  and  all 
metals  except  gold  and  silver.  By  a  late  law  of  Massachusetts,  the 
cwt.  contains  100  Ibs.  instead  of  112  Ibs. 

A  ton  is  reckoned  at  the  custom-houses  of  the  United  States  at 
2240  Ibs. 

LONG    MEASURE. 

3  Barleycorns,  or  12  Lines  make     1  Inch, 

12  Inches  " 

3  Feet  " 

6  Feet  " 

54  Yards,  or  16$  Feet  " 

40  Rods  " 

8  Furlongs  " 

3  Miles  " 

694  Miles  nearly  " 

360  Degrees  " 

in.  ft. 

12  =,  1  yd. 

36       =  3       =  1 

198      =  164     =  54 


7920       = 
63360      = 


660 
5280 


1760 


1  Inch,           marked 

in. 

Foot,               " 

ft. 

Yard, 
Fathom,           " 

yd. 
fth. 

Rod,  or  Pole,  " 

rd. 

Furlong,           " 

fur. 

Mile, 

m. 

League 

lea. 

Degree,           "       Deg. 
Circle  of  the  Earth. 

or0 

rd. 

=          1                 fur. 

=       40       =       1 

m. 

=     320      =       8      == 

1 

46 


MEASURES. 


[SECT.  ix. 


CLOTH    MEASURE. 


24  Inches 

make 

Nail                      marked    na. 

4     Nails 

u 

Quarter  of  a  yard,     "          qr. 

4    Quarters 

« 

Yard,                         "        yd. 

3     Quarters 

ii 

Ell  Flemish,             "    E.  F. 

5     Quarters 

u 

Ell  English, 

4     Quarters  1|  inch 

« 

Ell  Scotch,                "    E.  S. 

SQUARE    ME 

iSURE. 

144     Square  inches 

make 

Square  foot,         marked      ft. 

9     Square  feet 

H 

Square  yard,             "         yd. 

304  Square  yards                     " 
2724  Square  feet                       " 
40     Square  rods  or  poles        " 

Square  rod  or  pole,    '            p. 
Square  rod  or  pole,    "          p. 
Rood,                          "         R. 

4     Roods 

11 

Acre,                          "         A. 

640     Acres 

« 

Square  mile,              "    S.  M. 

in.                        ft. 

144=.                1 

yd. 

1596  =                9 

=                 1                    P. 

39204  =          2724 

=            304  =               1                R. 

1568160  =        10890 

=        1210  =          40  =        1          A. 

6272640  =        43560 

=        4840  =         160  =        4=1   s.M. 

4014489600  =  27878400 

=  3097600  =  102400  =  2560  =  640  =  1 

DRY    MEASURE. 

2  Pints 

make        Quart,               marked        qt. 

4  Quarts 

"           Gallon,                   "           gal. 

2  Gallons 

"           Peck,                   .  "            pk. 

4  Pecks 

««           Bushel,                   "            bu. 

36  Bushels 

"           Chaldron,               "            ch. 

pts.                               gal 

8          =              1 

pk. 

16          =              2 

=                  1                         bu. 

64          =              8 

=               4=1                       ch. 

2304          =          288 

=           144         =         36           =1 

NOTE. — This  measure  is  applied  to  all  goods  that  are  not  liquid  and 
are  sold  by  measure,  as  corn,  fruit,  salt,  coals,  &c.  A  Winchester  Bushel 
is  18£  inches  in  diameter,  and  8  inches  deep.  The  standard  Gallon  Dry 
Measure  contains  268|  cubic  inches. 


2  Pints 

4  Quarts 
32  Gallons 
54  Gallons 

2  Hogsheads 

2  Butts 


ALE   AND   BEER    MEASURE. 


make 


Quart, 
Gallon, 
Barrel, 
Hogshead, 
Butt, 
Tun, 

marked     qt. 

"         gal. 
bar. 
"       hhd. 
"       butt. 
"        tun. 

SECT.  IX.] 


MEASURES  AND  TIME. 


47 


pts. 

8 

256 
432 
864 


qt. 

4 

128 
216 
432 


32 
54 

108 


bar. 
=        1 

-       1» 
=      3 


hhd. 
1 


butt. 
=        1 


NOTE.  —  By  a  law  of  Massachusetts,  the  Barrel  for  cider  and  beer 
shall  contain  32  gallons,  but  in  some  other  States  it  is  of  different  capa- 
city. The  Ale  Gallon  contains  282  cubic  or  solid  inches. 

Milk  is  sold  by  the  Beer  Gallon. 


WINE    MEASURE. 


4  Gills 

2  Pints 

4  Quarts 

42  Gallons 

63  Gallons,  or  1£  Tierces 

2  Tierces 

2  Hogsheads 

2  Pipes,  or  4  Hhds. 


ptS. 

2 
8 

336 

504 

672 

1008 

2016 


qt. 

4 

168 
252 
336 
504 
1008 


1 

42 
63 

84 
126 
252 


make 


Pint, 
Quart, 
Gallon, 
Tierce, 
Hogshead, 
Puncheon, 
Pipe  or  Butt, 
Tun, 

marked    pt. 
"          qt. 
"       gal. 
"      tier. 
"      hhd. 
"      pun. 
"         pi. 
"      tun. 

tier. 


=     2 


hhd. 
1 

1J  = 
3  =  2  = 
6  =  4  = 


pun. 
1 


tun. 
1 


NOTE.  —  The  Wine  Gallon  contains  231  cubic  inches.  Water,  wine, 
and  spirits  are  measured  and  sold  by  this  measure. 

A  cubic  foot  of  distilled  water  weighs  158  ounces  Avoirdupois. 

The  English  Imperial  Gallon  contains  277£  cubic  inches,  and  weighs 
10  Ib.  Avoirdupois,  or  12  Ib.  1  oz  16  dwt.  16gr.  Troy.  There  is  no  legal 
measure  in  the  United  States  for  tierce,  hogshead,  puncheon,  pipe,  or  butt. 


60  Seconds,  or  60" 
60  Minutes 
24  Hours 

7  Days 

4  Weeks 
13  Months,  1  day,  6  hours,  or 

365  days,  6  hours, 
12  Calendar  months 


OF    TIME. 

make 

Minute, 

(4 

Hour, 

(« 

Day, 

(4 

Week, 

« 

Month, 

'  or  I     1  Julian  Year, 

"       1  Year, 

marked 


m. 
h. 
d. 
w. 
mo. 

y. 
y- 


48  MOTION  AND  DISTANCES.  [SECT.  ix. 

sec.  m. 

60  =  1  h. 

3600  =  60  =  1                 d. 

86400  =  1440  =  24  =         1             w. 

604800  =  10080  =  168  =         7=1          mo. 

2419200  =  40320  =  672  =       28       =    4     =     1          y. 

31557600  =  525960  =  8766  =    365|                  =               1 

NOTE.  — The  true  solar  year  is  the  time  measured  from  the  sun's  leav- 
ing either  equinox  or  solstice,  to  its  return  to  the  same  again.  A  period- 
ical year  is  the  time  in  which  the  earth  revolves  round  the  sun,  and  is 
365  d.  6  h.  9  m.  14£  sec.,  and  is  often  called  the  Sidereal  year.  The  civil 
year  is  that  which  is  in  common  use  among  the  different  nations  of  the 
world,  and  contains  365  days  for  three  years  in  succession,  but  every  fourth 
year  contains  366  days.  When  any  year  can  be  divided  by  four,  without 
any  remainder,  it  is  leap  year,  and  has  366  days. 

w.         d.          h.  mo.         d.          h. 

Or,    52        1        6      =5      13        1        6      =      1  Julian  Year. 

d.  h.  m.  sec. 

But,     365  5  48  57       =       1  Solar  Year. 

And,    365  6'  9  144=1  Sidereal  Year. 

The  days  in  each  month  are  as  follows :  —  January,  March,  May, 
July,  August,  October,  and  December  have  31  days  each  ;  April, 
June,  September,  and  November  have  30  days  each ;  February  has 
28  days,  excepting  leap  year,  when  it  has  29. 

CIRCULAR   MOTION. 

60  Seconds,  or  60 "                 make     1  Prime  minute,        marked  / 

60  Minutes                                  ««  '      1  Degree,                        "  ° 

30  Degrees                                  «        1  Sign,                            «  s. 

12  Signs,  or  360  Degrees,  the  whole  great  circle  of  the  zodiac. 

6'6  =  1 

3600  =  60  =  1                      8 

108000  =  1800  =  30  =           1 

1296000  =  21600  =  360  =         12         =        zodiac. 

MEASURING   DISTANCES. 

7$  Inches  make  1  Link. 

25  Links  "  1  Pole. 

100  Links  «  1  Chain. 

10  Chains  "  1  Furlong. 

8  Furlongs  "  1  Mile. 


SECT.  IX.] 


MISCELLANEOUS  TABLE. 


49 


Inches. 

Link. 

*ll 

0  = 

1 

Pole. 

192 

= 

25 

__ 

1 

Chain. 

792 

— 

100 

_. 

4 

=             1 

Furlong. 

7920 



1000 

— 

40 

=         10 

=          1 

Mile. 

63360 

= 

8000 

= 

320 

=        80 

==        8      = 

1 

SOLID   MEASURE. 


1728  Inches 
27  Feet 

40  Feet  of  round  timber 
128  Feet,  i.  e.  8  in  length,  4  in  breadth,  > 
and  4  in  height,  f 


make 


1  Foot. 
1  Yard. 
1  Ton. 

1  Cord  of  wood. 


NOTE.  —  One  ton  of  round  timber,  as  usually  surveyed,  contains 
solid  feet. 


MISCELLANEOUS  TABLE. 


A  gallon  of  train  oil 
A  stone  of  butcher's  meat 
A  gallon  of  molasses 
A  stone  of  iron 
A  tod 

A  firkin  of  butter 
A  firkin  of  soap 
A  quintal  of  fish 
A  weigh 
A  sack 

A  puncheon  of  foreign  prunes 
A  last 

A  fother  of  lead 
A  barrel  of  anchovies 
"          raisins 
flour 

pork  or  beef 
soap 

"  shad  or  salmon  in  Connect-  ) 

icut  or  New  York         £ 
fish  in  Massachusetts 
cider  and  beer 
herrings  in  England 
"  salmon  or  eels  do. 

8  bushels  of  salt,  measured  on 
board  the  vessel, 
7i     do.     measured  on  shore, 
5 


weighs 

M 


s 


7A  pounds. 

8  " 

11  " 

14  " 

28 

56 

94 

100 

182 

364 

1120  « 

4368  " 

19£  cwt. 

30  pounds. 

112  " 

196  " 

200  " 

256  " 

200  " 

30  gallons. 

32  " 

32  " 

42  " 

1  hogshead. 


50  COMPOUND  ADDITION.  [SECT.  x. 


3  hoops  make 

40  casts 
10  hundred 
12  units,  or  things, 
12  dozen 
144  dozen  " 


cast, 
hundred, 
thousand, 
dozen. 


great  gross. 


SECTION  X. 
COMPOUND  ADDITION. 

WHEN  numbers  are  applied  to  things,  the  measure  or  value 
of  which  is  expressed  by  different  denominations,  they  lose  their 
abstract  character,  and  become  subject  to  restrictions,  imposed 
upon  them  by  the  denomination  to  which  they  are  applied. 
Thus,  when  we  say  six  cents,  ten  days,  or  three  inches,  we  have 
not  only  the  idea  of  number,  but  also  the  idea  of  a  certain 
value  or  measure,  which  subjects  the  number  in  connection  with 
it  to  certain  limitations.  And,  when  used  in  such  connections, 
we  call  numbers  denominate.  Thus  in  §£.  4s.  7d.  the  num- 
bers 6,  4,  and  7  are  denominate  numbers,  so  called,  because 
they  are  applied  to  express  each  a  particular  denomination. 

When  now  we  have  several  numbers  of  different  denomina- 
tions, which  we  wish  to  add  together,  we  call  the  process  by 
which  this  is  done  Compound  Addition  ;  which  we  define  by 
saying,  ^ 

That  it  consists  in  adding  together  two  or  more  numbers  of 
different  denominations  to  find  the  sum  total. 

RULE. 

Write  all  the  given  numbers  of  the  same  denomination  under  each 
other ;  as  dollars  under  dollars,  cents  under  cents,  <5fC.  Then  add  to- 
gether  the  numbers  of  the  lowest  denomination  and  divide  the  sum  by  the 
number  which  it  takes  of  that  denomination  to  make  one  of  the  denom- 
ination next  above  it,  and  set  the  remainder  directly  under  the  column 
that  has  been  added.  Carry  the  quotient  to  the  column  of  the  next  de- 
nomination, and  add  as  before,  dividing  by  the  number  which  it  takes  of 
this  denomination  to  make  one  of  the  denomination  next  above  it,  setting 
down  the  remainder  and  carrying  the  quotient  as  before,  and  thus  pro- 
ceed till  the  column  of  the  highest  denomination  is  added,  under  which 
place  its  whole  sum,  and  the  numbers  expressing  the  several  denomina- 
tions will  be  the  sum  total  required. 


SECT,  x.]  COMPOUND  ADDITION.  51 

EXAMPLES. 


i. 

UNITED 

STATES  MONEY. 
2. 

3. 

ft. 

cts. 

m. 

ft. 

cts. 

m. 

E. 

ft.  cts. 

m. 

325 

67 

3 

28 

15 

6 

71 

3 

41 

5 

186 

35 

8 

16 

16 

3 

61 

6 

82 

6 

161 

89 

9 

63 

81 

5 

16 

1 

96 

2 

987 

15 

8 

14 

61 

6 

41 

7 

82 

1 

891 

61 

6 

38 

74 

5 

54 

8 

36 

3 

176 

81 

3 

16 

16 

8 

41 

9 

48 

5 

2729 

51 

7 

ENGLISH  MONEY. 

4. 

5. 

6. 

£. 

8. 

d. 

£. 

s. 

d. 

£. 

s. 

d. 

qr. 

471 

16 

9 

28 

6 

9^ 

31 

17 

9 

2 

147 

17 

8 

15 

16 

ni 

16 

16 

6 

1 

613 

13 

11 

31 

13 

HJfc 

16 

11 

11 

1 

115 

11 

7 

14 

16 

9 

19 

19 

9 

3 

41 

19 

6 

17 

17 

7| 

61 

17 

1 

3 

48 

12 

2 

32 

18 

8* 

14 

14 

4 

2 

1439 

11 

7 

TROY  WEIGHT. 

7. 

8. 

Ib. 

oz. 

dwt. 

&• 

Ib. 

oz. 

dwt. 

gr. 

16 

11 

19 

23 

123 

9 

7 

13 

31 

10 

18 

16 

98 

11 

17 

14 

63 

9 

12 

15 

49 

7 

13 

21 

17 

8 

13 

12 

13 

10 

10 

20 

61 

7 

12 

16 

47 

9 

19 

23 

17 

6 

17 

22 

51 

5 

15 

15 

209 

7 

15 

8 

APOTHECARIES' 

WEIGHT. 

9.                             10. 

Ib 

§ 

3 

B 

gr. 

ft 

§  3 

9 

gr. 

27 

11 

7 

2 

19 

37 

9  6 

1 

18 

16 

10 

6 

1 

13 

14 

4  4 

2 

11 

41 

9 

3 

2 

16 

61 

6  3 

2 

6 

38 

10 

5 

2 

14 

41 

4  7 

2 

16 

41 

4 

4 

1 

11 

39 

8  4 

1 

12 

16 

6 

6 

2 

6 

51 

11  7 

2 

19 

183 

6 

3 

1 

19 

52 


COMPOUND  ADDITION. 


[SECT.  x. 


AVOIRDUPOIS   WEIGHT. 


11. 


Ton.  cwt.  qr.  Ib.  oz.  dr. 

61  19  3  27  15  15 

63  13  3  16  11  11 

51  12  3  17  7  6 

61  16  1  11  12  12 

13  13  3  12  13  15 

71  18  2  13  14  14 

324  15  2  16  12  9 


12. 

cwt.  qr.  Ib.  oz.  dr. 

61  2  11  11  14 

16  3  15  15  11 

41  3  13  9  9 
38  2  11  10  10 

42  1  9  8  13 
31  3  27  11  12 


LONG   MEASURE. 


13. 

14. 

Deg. 

m. 

fur. 

rd. 

ft. 

in. 

br. 

m. 

fur. 

rd.  yd. 

ft. 

In.  br. 

17 

69 

7 

39 

16 

11 

2 

69 

7 

31  5 

2 

11   1 

61 

62 

3 

17 

12 

9 

1 

16 

6 

16  4 

1 

6  2 

16 

16 

6 

16 

13 

10 

2 

61 

7 

32  3 

2 

10  1 

48 

19 

3 

15 

15 

6 

1 

73 

3 

16  4 

2 

9  2 

17 

58 

6 

33 

14 

7 

1 

19 

4 

14  1 

1 

8  2 

33 

35 

5 

19 

9 

9 

2 

75 

5 

25  5 

2 

7  1 

195 

54j 

h  1 

24 

j 

r   7 

0 

T 

.=4 

•i 

h=6 

195    54      5    24 


1     0 


NOTE. —  As  half  a  mile  is  equal  to  4  furlongs,  we  add  them  to  the  1 
furlong,  which  make  5  furlongs.  And  as  half  a  foot  is  equal  to  6  inches, 
we  add  them  to  the  7  inches,  which  make  13  inches  ;  and  these  are  equal 
to  1  foot  1  inch.  By  the  same  method,  we  obtain  the  result  in  the  14th 
and  17th  questions. 


CLOTH   MEASURE. 


15. 

yd.  qr.  na.  in. 

37  3  3  2 

61  3  1  1 

13  2  2  2 

32  1  1  1 

61  2  2  2 

22  1  3  0 

229  3  3  1 


16. 

E.E.  qr.  na.  in. 

671  1  1  1 

161  3  3  2 

617  3  1  2 

178  321 

717  2  1'  2 

166  3  2  1 


SECT,  x.]  COMPOUND  ADDITION.  53 


LAND  OR  SQUARE  MEASURE. 
17. 

18. 

A. 

R. 

p.    ft.     in. 

A.   R.   p. 

ft. 

761 

3 

37  260  125 

38  1 

39 

272 

131 

2 

16  135  112 

61  3 

38 

167 

613 

1 

14  116  131 

35  3 

19 

198 

161 

3 

13  116  123 

47  3 

16 

271 

321 

2 

31   97   96 

86  2 

13 

198 

47 

3 

19   91   48 

46  1 

14 

269 

2038 

1 

13    2   95 

SOLID 

MEASURE. 

19. 

20. 

Ton. 

ft. 

in. 

Cord. 

ft. 

in. 

29 

36 

1279 

61 

127 

1161 

69 

19 

1345 

37 

89 

1711 

67 

18 

1099 

61 

98 

1336 

71 

14 

1727 

43 

56 

1678 

43 

35 

916 

91 

119 

1357 

53 

17 

1719 

81 

115 

1129 

335 

23 

1173 

WINE 

MEASURE. 

21. 

22. 

Tun. 

61 

hhd. 

3 

gal.  qt.  pt. 

62  3  1 

hhd. 

67 

fi 

qt.  pt. 

O    J. 

39 

2 

16  1  1 

16 

16 

3  0 

68 

3 

57  2  1 

39 

16 

3  0 

87 

3 

45  3  1 

47 

62 

1  1 

47 

2 

59  3  1 

43 

57 

3  0 

47 

3 

39  2  1 

71 

61 

3  1 

354 

0 

30  1  0 

- 

ALE  AND 

BEER  MEASURE. 

23. 

24. 

Tun. 

hhd. 

gal.   qt. 

hhd. 

gal. 

qt.  pt. 

46 

3 

50  3 

161 

c  o 
Do 

3  1 

91 

2 

48  3 

371 

52 

3  1 

17 

3 

18  0 

98 

19 

1  0 

81 

3 

38  2 

47 

43 

1  0 

41 

1 

47  1 

61 

43 

1  1 

37 

2 

29  3 

42 

27 

3  1 

317 

2 

17  0 

5* 

54  COMPOUND  ADDITION.  [SECT.  x. 

DRY   MEASURE. 


bu. 

37 

25. 

f 

qt.  pt. 

5  1 

ch. 

16 

26. 
bu. 

31 

t 

I1' 

61 

3 

7 

1 

39 

31 

3 

1 

32 

2 

2 

0 

14 

16 

3 

1 

71 

1 

6 

1 

55 

15 

3 

0 

61 

1 

3 

1 

71 

17 

3 

1 

32 

3 

3 

1 

42 

14 

3 

1 

298 

0 

4 

1 

TIME. 

27. 

2 

3. 

y. 

mo. 

w. 

d, 

h. 

m. 

8. 

y. 

mo. 

d. 

h. 

57 

11 

3 

6 

23 

29 

55 

13 

5 

29 

17 

31 

11 

1 

3 

19 

19 

39 

61 

11 

17 

21 

46 

9 

2 

2 

17 

28 

56 

15 

9 

19 

16 

43 

10 

1 

1 

18 

17 

48 

61 

10 

25 

23 

32 

9 

1 

3  16 

23 

28 

41 

4 

16 

17 

14 

1 

1 

5  22 

28 

16 

18 

5 

9 

6 

227 

2 

0 

3 

21 

28 

2 

CIRCULAR 

MOTION. 

29. 

30. 

S. 

O 

i 

it 

s. 

o 

. 

• 

4 

29 

59 

59 

11 

11 

16 

51 

6 

17 

17 

29 

6 

6 

6 

16 

11 

16 

56 

58 

9 

14 

56 

56 

9 

13 

46 

51 

3 

29 

29 

49 

5 

27 

16 

42 

9 

17 

18 

58 

% 

25 

17 

17 

6 

13 

13 

52 

5 

10 

35 

16 

NOTE.  —  We  divide  the  sura  of  the  signs,  in  these  questions,  by  12,  and 
write  down  the  remainder,  because  it  is  Circular  Motion. 

MEASURING   DISTANCES. 

31.  32. 

m.  fur.  ch.  p.   1.  m.  fur.  ch.  p.   1. 

17  7  9  3  24  27  4  3  1  21 

16  3  4  1  15           29  3  1  3  23 
27  4  6  2  17           67  3  3  1  19 

18  6  3  3  21           21  7  1  3  16 
61  7  7  2  16           16  7  9  3  13 

17  1  8  2  19  31  4  8  1  20 


160  0  1  1  12 


SECT,  xi.]  COMPOUND  SUBTRACTION.  55 

SECTION  XI. 

COMPOUND  SUBTRACTION. 

COMPOUND  SUBTRACTION  teaches  to  find  the  difference  be- 
tween two  numbers  of  different  denominations. 

RULE. 

Write  the  smaller  compound  number  under  the  greater,  in  the  order 
of  the  different  denominations,  as  pounds  under  pounds,  shillings  under 
shillings,  <5fc.  Begin  with  the  lowest  denomination  and  subtract  each 
lower  number  from  the  one  above  it,  and  write  the  difference  underneath. 
If,  in  any  denomination,  the  lower  number  be  greater  than  the  one  above 
it,  add  to  the  upper  number  the  number  required  of  this  denomination 
to  make  one  of  the  next  higfier ;  and,  from  the  number  thus  obtained, 
subtract  the  lower  number  and  set  down  the  remainder  underneath. 
Carry  one  to  the  next  denomination  in  the  subtrahend,  and  proceed  in 
like  manner  with  the  subtraction,  till  the  operation  has  been  performed 
in  all  the  columns,  setting  down  the  entire  difference,  between  the  upper 
and  lower  numbers  of  the  highest  denomination,  and  the  result  will  be 
the  difference  required. 

NOTE.  — The  reason  for  increasing  the  number  of  the  minuend  by  the 
number  required  of  a  lower  denomination  to  makeone  of  the  next  higher, 
is  precisely  the  same  as  that  for  which  we  add  ten  to  a  figure  of  the 
minuend,  in  Simple  Subtraction.  In  both  cases  we  add  the  number  de- 
noting the  ratio  between  the  denomination  in  question,  and  the  next 
higher  number  of  the  minuend.  In  Simple  Subtraction,  the  ratio  of  in- 
crease from  right  to  left  being  uniformly  tenfold,  we  add  ten,  while  in  the 
case  of  farthings,  pence,  and  shillings,  we  add  4,  12,  and  20. 

EXAMPLES. 

UNITED    STATES    MONEY. 

1.  2. 

$        cts.      m.  $         cts.      m. 

169     81     3  681     16     7 

85     93     8  189    43    8 


83     87     5 


ENGLISH    MONEY. 


£.  s.  d.  £.  B. 

87  16  3J  617  11 

19  17  9£  181  15 

67  18  6j 


56  COMPOUND  SUBTRACTION.  [SECT.  xi. 


5. 

TROY   WEIGHT. 

6. 

lb 

oz. 

dwt.      CT.                                          lb. 

oz. 

dwt. 

gr 

71 

3 

12 

15                              58 

5 

12 

10 

16 

10 

17 

20                             19 

9 

17 

21 

54 

4 

14 

19 

APOTHECARIES'  WEIGHT. 

7. 

8. 

fe 
71 

S 

5    9   gr.                       ft     S 
3     1     14                       15    2 

2 

B 

0 

15 

18 

6 

7    2     19                        99 

1 

1 

18 

52 

6 

3 

1     14 

E 

AVOIRDUPOIS   WEIGHT. 

9.                                                             10. 

T. 

71 

cwt. 

18 

T 

lb.    oz.       dr.                         cwt. 

13     1     13                  73 

T 

lb. 

15 

oz. 
13 

19 

19 

2 

16    8      5                  19 

i 

19 

15 

51 

18 

2 

24    9      8 

CLOTH   MEASURE. 

11, 

12. 

yd.    qr. 

67     1 

na. 

1 

in. 

1 

E.E. 

51 

na. 

3 

18 

2 

2 

2 

19 

3 

1 

48 

2 

2 

i 

LONG   MEASURE. 

13. 

14. 

m. 

16 

fur. 
7 

rd. 

18 

ft. 

3 

in.  bar.           deg.      m.     fur. 

21          38    41     3 

rd. 

29 

yd. 

2 

ft.     in.  bar. 

172 

9 

7 

19 

16 

82          29     36     5 

31 

3 

1     9     1 

6 

7 

38 

2£ 

5    2 

£ 

=  6     0 

6 

7    38 

2 

11     2 

NOTE  —  As  half  a  foot  is  equal  to  6  inches,  we  add  them  to  the  5 
inches,  which  make  11  inches.  The  same  principle  is  adopted  in  the 
14th,  15th,  and  16th  examples. 

LAND    OR    SQUARE    MEASURE. 
15.  16. 

A.      R.       p.          ft.          In.  A.      R.       p.        yd.     ft.       in. 

56  1  19  119  110      13  1  15  19  1  17 
17  3  13  127  113      9  3  16  30  5  19 


38  2   5  264   33 


SECT.  XI.] 


COMPOUND  SUBTRACTION. 


57 


SOLID    MEASURE. 


17. 

18. 

Tons.      ft.          in. 

Corda.        ft.           in. 

49     13     1611 

361       47     1178 

18     15     1719 

197     121     1617 

30    37     1620 

WINE   MEASURE. 

19. 

20. 

Tun.  hhd.     gal.    qt. 

79     3     19     1 

pt.                                     hhd.    gal.    qt.    pt. 

1                           16     1     1     0 

11     1     28     2 

1                             9221 

68     1     53    3 

0 

ALE 

AND   BEER   MEASURE. 

21. 

22. 

Tun.    hhd.    gal.     qt. 

pt.                                         hhd.        gal.    qt. 

63     1     15     1 

0                             769     18     1 

19     3     16     3 

1                              191     19     3 

43     1     52     1 

1 

DRY  MEASURE. 

23. 

24. 

ch.        bu.    pk.    qt. 

56      2     1     1 

ch.        bu.    pk.    qt. 

39     12    2     1 

38       3     1     2 

12     25     3     5 

17     34    3     7 

TIME. 

25. 

26. 

mo. 

d.        h.        m.        s. 

y.        m.        w.     d.       h.       m.         s. 

6 

16     13    27     19 

48      0    2    5     19    27     31 

1 

22     16    41     37 

19     10    3    7    21     38    56 

4 

23    20    45    42 

CIRCULAR   MOTION. 

27. 

28. 

S.       o           in 

S.        o          i         ,, 

6     11     12    48 

4     19    41     22 

9      8     15    56 

1     22     19    28 

9      2     56     52 

58  COMPOUiND  ADDITION.  [SECT.  w. 

MEASURING    DISTANCES. 

29.  30. 

m.     fur.    ch.     p.        1.  m.     fur.    ch.  p.        1. 

21     1     3    2     19  28    6     1  2     18 

19    2     1     3    21  15    7     3  1     19 


1     7     1     2    23 

EXERCISES  IN  COMPOUND  ADDITION  AND 
SUBTRACTION. 

1.  What  is  the  sum  of  16^.  5s.  8d.  2qr.  —  3l£.  16s.  lid. 
3qr.—  2L£.  11s.   Iqr.  —  19c£.  Os.   lOd.  3qr.  —  13£.  13s.  7d. 
3qr.  and28<£.  17s.  5d.   Iqr.  ?  Ans.   13 1£.  5s.  8£d. 

2.  Bought  of  a  London  tailor  a  vest  for   !«£.  13s.  4d.,  a 
coat  for  7<£.  12s.  9d.,   pantaloons  for  2£.  3s.  9d.,  and  sur- 
tout  for  9<£.  8s.  Od. ;  what  was  the  whole  amount  ? 

Ans.  20JE.  17s.  lOd. 

3.  Bought  a  silver  tankard,  weighing  lib.  8oz.  17dwt.  14gr., 
a  silver  can,  weighing  lib.  2oz.   12dwt,  a  porringer,  weighing 
lloz.  19dwt.  20gr.,  and  three  dozen  of  spoons,  weighing  lib. 
9oz,  15dwt.   lOgr. ;  what  was  the  whole  weight  ? 

Ans.  51b.  9oz.  4dwt  20gr. 

4.  What  is  the  weight  of  a  mixture  of  31b  45  23  29  14gr. 
of' aloe,  2ft>  7§    63    IB   13gr.  of   picra,  and   lib  105   13 
29   17gr.  of  saffron  ?  Ans.  7tb    105   33    19  4gr. 

5.  Add   321b  95    13  29   14gr. ;  13  fo  75   63  19  13gr. ; 
and  16ft»  115   73    19   12gr.  together. 

Ans.  63fc  45   73  29   19gr. 

6.  Sold  4  loads  of  hay  ;  the  first  weighed  27cwt.  3qr.  181b. ; 
second,  31cwt.  Iqr.  151b.  ;  third,  19cwt.  Iqr.  151b. ;  and  fourth, 
38cwt.  2qr.  271b. ;  what  is  the  weight  of  the  whole  ? 

Ans.  117cwt.   Iqr.    191b. 

7.  Bought  5  pieces  of  broadcloth  ;  the  first  contained  17yd. 
3qr.  2na. ;  second,  13yd.  2qr.  Ina.  ;  the  third,  87yd.  Iqr.  3na. ; 
the  fourth,  27yd.  Iqr.  2na. ;  and  the  fifth,  29yd.   Iqr.  2na. ; 
what  was  the  whole  quantity  purchased  ? 

Ans.  175yd.  2qr.  2na. 

8.  A  pedestrian  travelled,  the  first  week,  371m.  3fur.  37rd. 
5yd.  2ft.  lOin. ;  the  second  week,  289m.  2fur.    18rd.   3yd.  1ft. 
9in.;  and  the  third  week  he  travelled  399m.  7fur.  3ft.  llin. ; 
how  many  miles  will  he  have  travelled  ? 

Ans.   1060m.  5fur.   16rd.  5yd.  1ft. 

9.  A  man  has  3  farms ;  the  first  contains  186A.  3R.  14p. ; 


SECT.  XL]  COMPOUND  SUBTRACTION.  59 

the  second,  286A.  17p. ;  and  the  third,  115A.  2R. ;  how  much 
do  they  all  contain  ?  Aris.  588 A.   IE,.  31p. 

10.  The  Moon  is  5s.  18°  14'   17"  east  of  the  Sun ;  Jupiter 
is  7s.  10°  29'  28"  east  of  the  Moon ;  Mars  is  11s.  12°   11' 
56"  east  of  Jupiter ;  and  Herschel  is  7s.    18o  38'  15"  east  of 
Mars  ;  how  far  is  Herschel  east  from  the  Sun  ? 

Ans.  7s.  29o  33'  56". 

11.  I  have  4  piles  of  wood;  the  first  contains  7  cords,  76ft. 
1671m. ;  the  second,   16c.  28ft.   56in.;   the  third,  29c.  127ft. 
1000  in.  ;  and  the  fourth,  29c.  10ft.  1216in. ;  how  much  is  there 
in  all?  Ans.  82c.  115ft.  487in. 

12.  A  vintner  sold  at  one  time  73hhd.  43gal.  3qt.  Ipt.  of 
wine  ;  at  another,  27hhd.  3gal. ;  at  another,  15hhd.  3qt.  Ipt. ; 
and  at  another,  161hhd.  and  2qt. ;  how  much  did  he  sell   in 
all  ?  l    Ans.  276hhd.  48gal.  Iqt. 

13.  A  man  has  3  sons  ;  the  first  is  14y.  3mo.  2w.  5d.  old  ; 
the  second  is  9y.  lOmo.  3w.  4d.   23h.    12m.    15sec.  ;  and  the 
third  is  2y.  Imo.  3w.  2d.  7m.  ;  what  is  the  sum  of  their  ages  ? 
and  how  much  older  is  the  first  than  the  second  ? 

Ans.  26y.  3mo.  Iw.  4d.  23h.  19m.  15sec. 
"       4y.  5mo.  3w.  Od.    Oh.  47m.  45sec. 

NOTE.  —  When  4  weeks  are  reckoned  as  a  month,  it  requires  13  months 
to  make  one  year. 

14.  I  have  73A.  of  land  ;  if  I  should  sell  5A.  3R.  Ip.  7ft., 
how  much  should  I  have  left  ?      Ans.  67A.  OR.  38p.  265£ft. 

15.  A.  owes  B.  100<£,  ;  what  will  remain  due  after  he  has 
paid  him  3s.  6^d.  ?  Ans.  99  £.  16s.  5£d. 

16.  It  is  about  25,000  miles  round  the  globe  ;  if  a  man  shall 
have  travelled  43m.  17rd.  9in.   how  much  will  remain  to  be 
travelled  ?  Ans.  24,956m.  7fur.  22rd.  15ft.  9in. 

17.  Bought  7  cords  of  wood  ;  and  2  cords  78ft.  having  been 
stolen,  how  much  remained  ?  Ans.  4c.  50ft. 

18.  I  have   15  yards  of  cloth ;  having  sold  3yd.  2qr.  Ina., 
what  remains  ?  Ans.   1  lyd.  Iqr.  3na. 

19.  If  a  wagon  loaded  with  hay  weighs  43cwt.  2qr.   181b., 
and  the  wagon  is  afterwards  found  to  weigh  9cwt.  3qr.  231b., 
what  is  the  weight  of  the  hay  ?  Ans.  33cwt.  2qr.  231b. 

20.  Bought  a  hogshead  of  wine,  and  by  an  accident  8gal. 
3qt.  Ipt.  leaked  out ;  what  remains  ?  Ans.  54gal.  Ipt. 

21.  I  had  10A.  3R.  lOp.  of  land  ;  and  I  have  sold  two  house- 
lots,  one  containing  1A.  2R.  13p.,  the  other,  2A.  2R.  5p.  ;  how 
much  have  I  remaining  ?  Ans.  6A.  2R.  32p. 

22.  The  Moon  moves   13°  10'  35''  in  a  solar  day,  and  the 


60  REDUCTION.  [SECT.  xn. 

Sun  59'  8"  20"' ;  now  supposing  them  both  to  start  from  the 
same  point  in  the  heavens,  how  far  will  the  Moon  have  gained 
on  the  Sun  in  24  hours  ?  Ans.  12°  11*  26"  40'". 

23.  A  farmer  raised  136bu.   of  wheat ;    if  he  sells  49bu. 
2pk.  7qt.  Ipt.,  how  much  has  he  remaining  ? 

Ans.  86bu.  Ipk.  Oqt.  Ipt. 

24.  If  from  a  stick  of  round  timber,  containing  2T.   18ft. 
1410in.,  there  be  taken  38ft.  1720in.,  how  much  will  be  left  ? 

Ans.  IT.  19ft.  1418in. 

25.  If  from  lib.  of  ipecacuanha  there  be  taken  at  one  time 
4S  23  13gr.,  and  at  another,  35  13  29  14gr.,  how  much  will 
be  left?  Ans.  4§  33  29  13gr. 

26.  A  brewer  has  in  one  cellar  18bbl.  3gal.  2qt.  of  beer, 
and  in  another,  13bbl.  Ip.  ;  what  is  the  whole  quantity,  and 
how  much  more  is  in  one  cellar  than  the  other  ? 

Ans.  31bbl.  3gal.  2qt.  Ipt. 
"         5bbl.  3gal.  Iqt.  Ipt. 

27.  If  from  $  100.00  there  be  paid  at  one  time  $  17.28,5,  at 
another  time  $  10.00,5,  and  at  another  $  37.15,  how  much  will 
remain?  Ans.  $35.56. 


SECTION  XII. 

REDUCTION. 

THE  object  of  Reduction  is  to  change  the  denomination  of 
numbers  without  altering  their  value.  It  consists  of  two  parts, 
Descending  and  Ascending.  The  former  is  performed  by  Mul- 
tiplication, the  latter  by  Division. 

"'Reduction  Descending  teaches  to  bring  numbers  of  .a  higher 
denomination  to  a  lower ;  as,  to  bring  pounds  into  shillings,  or 
tons  into  hundred  weights. 

Reduction  Ascending  teaches  to  bring  numbers  of  a  lower 
denomination  into  a  higher ;  as,  to  bring  farthings  into  pence, 
or  shillings  into  pounds. 

REDUCTION  DESCENDING. 

EXAMPLE. 

1.  In  48£.  12s.  7d.  2qr.,  how  many  farthings  ? 


SECT,  xii.]  REDUCTION.  61 

In  this  example,  we  multiply  the  48,£.  by 
20,  because  it  takes  20  shillings  to  make  a 
pound  ;  and  to  this  product  we  add  the  12s. 
in  the  question.  Then  we  multiply  by  12, 
because  it  takes  12  pence  to  make  one  shil- 
ling ;  and  to  the  product  we  add  the  7 
pence  in  the  question.  We  then  multiply  by 
4,  the  number  of  farthings  in  a  penny,  and  to 
the  product  we  add  the  2  farthings,  and  the 
work  is  done. 

From  the  above  example  and  illustration,  we  deduce  the 
following 

RULE. 

Multiply  the  highest  denomination  given  by  the  number  required  of 
the  next  lower  denomination  to  make  one  of  the  denomination  next  above 
it,  and  add  to  the  product  thus  obtained  the  corresponding  denomina- 
tion of  tfie  multiplicand.  Proceed  in  this  ivay,  till  the  reduction  is 
brought  to  the  denomination  required  by  the  question. 

NOTE  1. —  To  multiply  by  i,  we  divide  the  multiplicand  by  2,  and 
to  multiply  by  £,  we  divide  by  4. 

NOTE.  2.  —  The  answers  to  Reduction  Descending  will  be  found  in  the 
questions  of  Reduction  Ascending. 

2.  In  127<£.   15s.  8d.  how  many  farthings  ? 

3.  In  28£.   19s.  lid.  3qr.  how  many  farthings? 

4.  In  378£.  how  many  pence  ? 

5.  How  many  grains  in  281b.   lloz.   12dwt.   15gr.  troy  ? 

6.  In  171b.   12dwt.  troy,  how  many  pennyweights  ? 

7.  If  a  silver  tankard  weigh  31b.  lloz.,  how  many  grains  will 
it  be  ? 

8.  How  many  scruples  in  231b.  apothecaries'  weight  ? 

9.  If  a  load  of  hay  weigh  3T.  16cwt.  2qr.  181b.,  how  many 
ounces  will  it  be  ? 

10.  Required  the  number  of  drachms  in  a  hogshead  of  sugar, 
weighing  2T.  17cwt.  3qr.  161b.  15oz.   13dr. 

11.  In  57yd.   how  many  nails  ? 

12.  In  83947E.E.  4qr.  how  many  nails  ? 

13.  In  2263E.F.  2qr.  how  many  quarters  ? 

14.  How  many  feet  in  79  miles  ? 

15.  How  many  inches  in  396  furlongs  ? 

16.  How  many  inches  from  Haverhill  to  Boston,  the  distance 
being  30  miles  ? 

17.  How  many  barleycorns  will  it  take  to  reach  round  the 
world  ? 

6 


62  REDUCTION.  [SECT.  xn. 

18.  In   403m.  7fur.  35rd.  2yd.  Oft.  Oin.   Ibar.    how   many 
barleycorns  ? 

19.  In  413Le.  2m.  2fur.  38rd.    lyd.  Oft.  7in.  how  many 
inches  ? 

20.  In  144m.  Ifur.  8rd.   lyd.   1ft.  how  many  feet  ? 

21.  How  many  inches  in  1051yd.  2ft.  5in.  ? 

22.  In  3576fur.  12rd.  3yd.  how  many  yards  ? 

23.  How  many  square  feet  in  25  acres  ? 

24.  How  many  square  rods  in  365  square  miles  ? 

25.  The  surface  of  the    earth  contains   196563942   square 
miles.     What  would  it  be  in  square  inches  ? 

26.  Required  the  number  of  feet  in  10A.  3R.  38p.  6yd.  5ft. 
72in. 

27.  In  2R.  Op.  24yd.  3ft.  how  many  inches  ? 

28.  In  1A.  3R.  34p.  27yd.  4ft.  54in.  how  many  inches  ? 

29.  In  17  cords  of  wood,  how  many  inches  ? 

30.  In  19  tons  of  round  timber,  how  many  inches  ? 

31.  How  many  cubic  feet  of  wood  in  128  cords  ? 

32.  In  4899hhd.  4gal.  3qt.  how  many  quarts  ? 

33.  In  1224  tuns  Ip.  Ihhd.  19gal.  Iqt.  Opt.  Igi.  how  many  gills? 

34.  How  many  pints  in  790p.  Ohhd.  58gal.  Oqt.  Ipt.  ? 

35.  In  460  butts  Ihhd.  31  gal.  of  beer,  how  many  gallons  ? 

36.  In  36hhd.  26gal.  3qt.  Ipt.  how  many  pints? 

37.  In  16  tons  of  round  timber,  how  many  inches  ? 

38.  How  many  seconds  from  the  deluge,  it  being  2348  years 
B.  C.,  to  the  year  1836  ? 

39.  How  many  days  did  the  last  war  continue,  it  having  com- 
menced June  18,  1812,  and  ended  Feb.  17,  1815? 

40.  How  many  pecks  in  676  chaldrons  ? 

41.  In  657  cents,  how  many  mills? 

42.  In  3165  dimes,  how  many  mills  ? 

43.  In  63  dollars,  how  many  cents  ? 

44.  In  27  eagles,  how  many  mills  ? 

REDUCTION  ASCENDING. 

EXAMPLE. 

1.  In  76789  farthings,  how  many  pounds  ? 

OPERATION.  Ans.  79<£.  19s.  9±d. 

4)76789  We  first  divide  by  4,  because  4  far- 

12)19197rTqr  things  make  a  penny.     We  then   di- 

OHM^QQ  QJ  '  vide  by  12'  because  12  Pence  make  a 

20)1599  9d. shilling.     Lastly  we  divide  by  20,  the 

79£.  19s.  9£d.      number  of  shillings  in  a  pound. 


SECT,  xii.]  REDUCTION.  63 

From  the  preceding  illustration  and  example,  we  deduce  the 


following 


RULE. 


Divide  the  lowest  denomination  given  by  the  number  which  it  takes  of 
that  denomination  to  make  one  of  the  next  higher ;  so  proceed,  until  it  is 
brought  to  the  denomination  required.  Any  remainders  occurring  in 
the  successive  divisions  will  be  of  the  same  denominations  with  the  divi- 
dends to  which  they  respectively  belong. 

NOTE  1.  —  To  divide  by  5^,  we  multiply  the  multiplicand  by  2,  and 
divide  the  product  by  11  ;  to  divide  by  16^,  we  multiply  by  2  and  divide 
by  33 ;  and  to  divide  by  272^,  we  multiply  by  4  and  divide  by  1089. 

NOTE  2.  —  The  answers  to  Reduction  Ascending  are  the  questions  in 
Reduction  Descending. 

NOTE  3.  —  When  we  divide  by  11  or  33,  and  there  is  a  remainder,  we 
divide  that  remainder  by  2  to  get  the  true  remainder;  and,  when  we  di- 
vide by  1081),  we  must  divide  the  remainder  by  4. 

2.  In  122672  farthings,  how  many  pounds  ? 

3.  In  27839  farthings,  how  many  pounds  ? 

4.  In  90720  pence,  how  many  pounds  ? 

5.  In  166S63  grains,  how  many  pounds  troy  ? 

6.  How  many  pounds  in  4092  pennyweights  ? 

7.  How  many  pounds  troy  in  22560  grains  ? 

8.  In  6624  scruples,  how  many  pounds  ? 

9.  In  137376  ounces,  how  many  tons  ? 

10.  In  1660157  drachms,  how  many  tons  ? 

11.  How  many  yards  in  912  nails  ? 

12.  Required  the  ells  English  in  1678956  nails. 

13.  Required  the  ells  Flemish  in  6791  quarters. 

14.  Required  the  miles  in  417120  feet. 

15.  Required  the  furlongs  in  3136320  inches. 

16.  Required  the  miles  in  1900800  inches. 

17.  How  many  degrees  in  4755801600  barleycorns  ? 

18.  How  many  miles  in  76789567  barleycorns  ? 

19.  How  many  leagues  in  78653167  inches  ? 

20.  Required  the  miles  in  761116  feet. 

2 1 .  Required  the  yards  in  37865  inches. 

22.  Required  the  furlongs  in  786789  yards. 

23.  How  many  acres  in  1089000  square  feet  ? 

24.  How  many  square  miles  in  37376000  square  rods  ? 

25.  How  many  square  miles   in  789,103,900,894,003,200 
square  inches  ? 

26.  In  478675  square  feet,  how  many  acres  ? 

27.  In  3167856  square  inches,  how  many  roods  ? 

28.  How  many  acres  in  12345678  square  inches  ? 

29.  How  many  cords  in  3760128  cubic  inches  ? 


64  REDUCTION.  [SECT.  in. 

30.  How    many   tons   of  round   timber  in    1313280   cubic 
inches  ? 

31.  How  many  cords  in  16384  cubic  feet? 

32.  How  many  hogsheads  of  wine  in  1234567  quarts  ? 

33.  How  many  tuns  of  wine  in  9877001  gills  ? 

34.  In  796785  pints  of  wine,  how  many  pipes  ? 

35.  How  many  butts  of  beer  in  49765  gallons  ? 

36.  In  15767  pints,  how  many  hogsheads  of  beer  ? 

37.  In  1105920  inches,  how  many  tons  of  round  timber  ? 

38.  In  132005440800  seconds,  how  many  years  ? 

39.  In  974  days,  how  many  years  and  calendar  months  ? 

40.  How  many  chaldrons  in  97344  pecks  ? 

41.  How  many  cents  in  6570  mills  ? 

42.  How  many  dimes  in  316500  mills  ? 

43.  How  many  dollars  in  6300  cents  ? 

44.  How  many  eagles  in  270000  mills  ? 

COMPOUND  REDUCTION. 

1.  In  57£.  15s.  how  many  dollars  ?        Ans.  $  192.50cts. 

2.  In  67 £.  14s.  9d.  how  many  crowns  at  6s.  7d.  each  ? 

Ans.  205cr.  5s.  2d. 

3.  How  many  pounds  and  shillings  in  678  dollars  ? 

Ans.  203£.  8s. 

4.  How  many  ells  English  in  761  yards  ? 

Ans.  608E.  E.  4qr. 

5.  How  many  yards  in  61  ells  Flemish  ?      Ans.  45yd.  3qr. 

6.  How  many  bottles,  that  contain  3  pints  each,  will  it  take 
to  hold  a  hogshead  of  wine  ?  Ans.   168. 

7.  How  many  steps,  2ft.  Sin.  each,  will  a  man  take  in  walk- 
ing from  Bradford  to  Newburyport,  the  distance  being  15  miles  ? 

Ans.  29700. 

8.  How  many  spoons,  each  weighing  2oz.   12dwt.,  can  be 
made  from  51b.  2oz.  Sdwt.  of  silver  ?  Ans.  24. 

9.  How  many  times  will  the  wheel  of  a  coach  revolve,  whose 
circumference  is  14ft.  9in.  in  passing  from  Boston  to  Washing- 
ton, the  distance  being  436  miles  ?  Ans.  156073fV9T. 

10.  I  have  a  field  of  corn,  consisting  of  123  rows,  and  each 
row  contains  78  hills,  and  each  hill  has  4  ears  of  corn  ;  now  if 
it  take  8  ears  of  corn  to  make  a  quart,  how  many  bushels  does 
the  field  contain  ?  Ans.   149bu.  3pk.  5qt.  Opt. 

11.  If  it  take  5yd.  2qr.  3na.  to  make  a  suit  of  clothes,  how 
many  suits  can  be  made  from  182  yards  ?  Ans.  32. 

12.  A  goldsmith  wishes  to  make  a  number  of  rings,  each 


SECT,  xii.]  REDUCTION.  65 

weighing  5dwt.  lOgr.,  from  31b.   loz.  2dwt.  2gr.  of  gold  ;  how 
many  will  there  be  ?  Ans.   137. 

13.  How  many  shingles  will  it  take  to  cover  the  roof  of  a 
building,  which  is  60  feet  long  and  56  feet  wide,  allowing  each 
shingle  to  be  4  inches  wide  and  18  inches  long,  and  to  lay  one 
third  to  the  weather?  Ans.  20160. 

14.  There  is  a  house  56  feet  long,  and  each  of  the  two  sides 
of  the  roof  is  25  feet  wide ;  how  many  shingles  will  it  take  to 
cover  it,  if  it  require  6  shingles  to  cover  a  square  foot  ? 

Ans.  16800. 

15.  If  a  man  can  travel  22m.  3fur.  17rd.  a  day,  how  long 
would  it  take  him  to  walk  round  the  globe,  the  distance  being 
about  25000  miles  ?  Ans.  1114ffff  days. 

16.  If  a  family  consume  71b.  lOoz.  of  sugar  in  a  week,  how 
long  would  lOcwt.  3qr.  161b.  last  them  ?        Ans.  160  weeks. 

17.  Sold  3  tons  17cwt.  3qr.  181b.  of  lead  at  7d.  a  pound  ; 
what  did  the  lead  amount  to  ?  Ans.  254c£.   10s.  2d. 

18.  What  will  5cwt.   Iqr.  lOlb.  of  tobacco  cost,  at  4£d.  a 
pound?  Ans.  ll£.  4s.  3d. 

19.  What  will  7  hogsheads  of  wine  cost,  at  9  cents  a  quart  ? 

Ans.  $  158.76. 

20.  What  will  15  hogsheads  of  beer  cost,  at  3  cents  a  pint  ? 

Ans.  $  194.40. 

21.  What  will  73  bushels  of  meal  cost,  at  2  cents  a  quart  ? 

Ans.  $  46.72. 

22.  A  merchant  has  29  bales  of  cotton  cloth  ;  each  bale  con- 
tains 57  yards  ;  what  is  the  value  of  the  whole  at  15  cents  a 
yard  ?  Ans.  $  247.95. 

23.  A  merchant  bought  4  bales  of  cotton  ;  the  first  contained 
6cwt.  2qr.  lllb. ;  the  second,  5cwt.  3qr.  161b. ;  the  third,  7cwt. 
Oqr.   71b.  ;  the  fourth,  3cwt.    Iqr.   171b.     He  sold  the    whole 
at  15  cents  a  pound  ;  what  did  it  amount  to  ?    Ans.  $  385.65. 

24.  A  merchant  having  purchased  12cwt.  of  sugar,  sold  at 
one  time   3cwt.  2qr.  lllb.  and   at  another  time  he  sold  4cwt. 
Iqr.   151b.  ;    what   is  the   remainder  worth,   at  15   cents  per 
pound  ?  Ans.  $  67.50. 

25.  Bought  4  chests  of  hyson  tea ;  the  weight  of  the  first  was 
2cwt.  Iqr.  71b.  ;  the  second,  3cwt.  2qr.  151b. ;  ihe  third,  2cwt. 
Oqr.  201b. ;  the  fourth,  5cwt.  3qr.  171b.  ;  what  is  the  value  of 
the  whole,  at  37£  cents  a  pound  ?  Ans.  $  589. 12^. 

26.  Purchased  a  cargo  of  molasses,  consisting  of  87  hogs- 
heads ;  what  is  the  value  of  it,  at  33  cents  a  gallon  ? 

Ans.  $  1808.73. 
6* 


66  REDUCTION.  [SECT.  XH. 

27.  From  a  hogshead  of  wine,  lOgal.  Iqt.  Ipt.  3  gills  leaked 
out.     The  remainder  was  sold  at  6  cents  a  gill ;  to  what  did  it 
amount  ?  Ans.  $  100.86. 

28.  A   man   has  3  farms ;  the  first  containing  100A.  3R. 
15rd. ;  the  second,   161  A.  2R.  28rd. ;  the  third,  360 A.  3R. 
5rd.     He  gave  his  oldest  son  a  farm  of  1 12A.  3R.  30rd. ;  his 
second  son  a  farm  of  316A.  1R.  18rd. ;  his  youngest  son  a 
farm  of  168 A.  3R.  13rd. ;  and  sold  the  remainder  of  his  land 
at  1  dollar  and  35  cents  a  rod  ;  to  what  did  it  amount  ? 

Ans.  $  5436.45. 

29.  A  grocer   bought   a   hogshead   of  molasses,  containing 
87gal.  Iqt.,  from  which   13  gallons  leaked  out ;  what  is  the 
remainder  worth,  at  1  cent  a  gill  ?  Ans.  $  23.76. 

30.  A  man  bought  4  loads  of  hay  ;  the  first  weighing  25cwt. 
Oqr.  171bs. ;  the  second,  37cwt.  2qr.   171b. ;  the  third,  18cwt. 
3qr.   141b.  ;  and  the  fourth,  37cwt.    Iqr.    171b. ;  what   is   the 
value  of  the  whole,  at  2  cents  a  pound  ?  Ans.  $266.74. 


REDUCTION  OF  THE  OLD  NEW  ENGLAND  CUR- 
RENCY  TO  UNITED  STATES  MONEY. 

The  original  currency  of  N.  E.  was  pounds,  shillings,  pence, 
and  farthings  ;  but,  on  the  adoption  of  the  Constitution  of  the 
United  States,  it  was  changed  to  dollars,  cents,  and  mills.  It  is 
frequently  necessary  to  reduce  the  former  to  the  present  cur- 
rency of  the  United  States ;  for  which  we  have  the  following 

RULE. 

If  pounds  only  are  given,  annex  three  ciphers  and  divide  by  3,  and 
the  quotient  will  be  the  sum  required  in  cents. 

If  pounds  and  an  even  number  of  shillings  are  given ,  annex  to  the 
pounds  half  the  number  of  shillings  and  two  ciphers,  and  divide  as  before. 

If  the  number  of  shillings  be  odd,  take  half  of  the  largest  even  number 
of  shillings  and  annex  it  to  the  pounds  with  the  figure  5  and  one  cipher , 
instead  of  two  as  above,  and  proceed  as  in  the  former  instances. 

If  pounds,  shillings,  pence,  and  farthings  are  given,  annex  to  the 
pounds  and  shillings,  as  before,  and  find  the  number  of  farthings  con- 
tained in  the  given  pence  and  farthings,  taking  care  to  increase  their 
number  by  I,  if  they  exceed  12,  and  by  2,  if  they  exceed  36.  Annex  the 
number  thus  obtained  to  the  pounds  in  such  a  way  that  the  units  of  the 
farthings  shall  occupy  the  third  place  from  the  pounds,  and  divide  by  3, 
as  before,  and  the  quotient  will  be  the  result  in  cents. 

NOTE.  — A  demonstration  of  this  rule  will  be  found  in  Sec.  XXIX. 


SECT.    XIII.] 


UNITED  STATES  MONE 


1.  Reduce  162<£. 

2.  Change  319=£. 

3.  Change  176<£, 

4.  Reduce  315  <£. 


EXAMPLES. 

to  United  States  Money 
3)162000 

$  540.00  Ans.  $  540. 

17s.  to  United  States  Money. 
3)319850 

$  1066:i6§         Ans.  $  1066.16f . 
17s.  8Jd.  to  United  States  Money. 
176850 
36 

3)176886 

f589^62  Ans.  $589.62. 

to  U.  S.  Money. 


5. 

6. 

7. 

8. 

9. 
10. 
11. 
12. 
13. 
14. 
15. 
16. 
17. 
18. 
19. 
20. 
21. 
22. 
23. 
24. 


SECTION  XIII. 

UNITED   STATES  MONEY. 

THE  denominations  of  Federal  or  United  States  Money  being 
in  the  ratio  of  10,  100,  and  1000  to  each  other,  operations  in- 


619  =£. 

to 

cc 

166  £. 

to 

cc 

318£. 

to 

cc 

101  £. 

to 

cc 

144  <£. 

to 

cc 

161  <£.  18s. 

to 

cc 

361^.  17s. 

to 

cc 

99  £.  11s. 

to 

cc 

100  <£.    9s. 

to 

cc 

661=£.    7s. 

to 

cc 

47  rf.  Us. 

to 

(C 

109  <£.    Is. 

to 

cc 

16  £.  17s.  6Jd. 

to 

cc 

69  £.    Is.  3Jd. 

to 

cc 

87,£.  16s.  lid. 

to 

cc 

14  £.    7s.  7^d. 

to 

cc 

73  £.    3s.  4£d. 

to 

cc 

47^.  12s.  lOd. 

to 

cc 

187  £.    5s.  Ofd. 

to 

cc 

10<£.    Os.  3|d. 

to 

cc 

Ans. 

$  1050.00 

cc 

2063.331 

cc 

553.33£ 

cc 

1060.00 

cc 

336.  66f 

cc 

480.00 

cc 

539.66| 

cc 

1206.  16§ 

cc 

331.83£ 

cc 

334.83£ 

cc 

2204.50 

cc 

158.50 

cc 

363.50 

cc 

56.25§ 

cc 

230.2  1£ 

cc 

292.82 

cc 

47.93§ 

cc 

243.89§ 

§1 

158.80f 

cc 

624.17| 

cc 

33.38 

68  UNITED   STATES  MONEY.  [SECT.  xm. 

volving  dollars,  cents,  and  mills  are  performed,  when  the  num- 
bers have  been  properly  set  down,  as  under  the  rules  for  sim- 
ple numbers. 

ADDITION. 

RULE. 

Write  dollars  under  dollars,  cents  under  cents,  and  mills  under  mills, 
and  then  proceed  as  in  Simple  Addition,  and  tfie  result  will  be  obtained 
in  the  several  denominations  added. 

NOTE.  —  In  all  operations  of  United  States  Money,  it  must  "be  borne  in 
mind  that  a  cent  is  one  hundredth  of  a  dollar,  and  hence,  in  arranging  a 
column  of  cents  or  annexing  any  number  of  cents  to  dollars,  1  cent,  2 
cents,  &c.,  must  be  written  .01,  .02,  &c.,  which  denote  one  hundredth, 
two  hundredth^,  &c. 

EXAMPLES. 

I.  2.  3.  4. 

ft    els.  m.  ft    cts.  m.  ft     cts.  m.  ft     cts.  m. 

375.87,5  78.19,3  171.01,3  861.07,3 

671.12.7  18.01,4  382.09,4  516.71,6 
387.14,3  91.03,8  999.90,0  344.67,3 
184.18,9  16.81,7  155.06,8  617.81,4 

147.75.8  81.47,6  48.15,3  169.97,3 
63.07,2              43.18,4                 49.61,9              810.42,6 


1829.16,4 

5.  Add  the  following  sums,  $18.16,5,  $701.63,  $  151.16,1, 
$  375.08,9,  and  $  471.01,7.  Ans.  $  1717.06,2. 

6.  Bought  a  horse  for  eighty-seven  dollars  nine  cents,  a  pair 
of  oxen  for  sixty-five  dollars  twenty  cents,  and  six   gallons  of 
molasses  for  two  dollars  six  cents  five  mills ;  what  was  the 
amount  of  my  bill  ?  Ans.  $  154.35,5. 

7.  Sold  a  calf  for  three  dollars  eight  cents,  a  bushel  of  corn 
for  ninety-seven  cents  five  mills,  and  three  bushels  of  rye  for 
three  dollars  five  cents  ;  what  was  the  amount  received  ? 

Ans.  $7.10,5. 

SUBTRACTION. 

RULE. 

Write  the  several  denominations  of  the  subtrahend  under  the  corre- 
sponding ones  of  the  minuend,  and  then  proceed  as  in  Simple  Subtraction, 
and  the  result  will  be  the  difference  in  the  several  denominations  sub- 
tracted. 


SECT,  xiii.]                        MULTIPLICATION.  69 

EXAMPLES. 

1.                                     2.                                     3.  4. 

$     cts.  m.                      $     eta.  m.                     9    eta.  m.  8     cts.  m 

871.16,1                478.47,7              167.16,3  163.16,7 

89.91,8                199.99,1               98.09,7  9.09,8 


781.24,3 

5.  Bought  a  farm  for  $  1728.90,  and  sold  it  for  $  3786.98, 
what  did  f  gain  by  my  bargain?  Ans.  $2058.08. 

6.  Gave  $  79.25  for  a  horse,  and  $  106.87,5  for  a  chaise,  and 
sold  them  both  for  $  200  ;  what  did  I  gain  ?     Ans.  $  13.87,5. 

7.  Bought  a  farm  for  $  8967,  and  sold  it  for  nine  thousand 
eight  hundred  seventy-six  dollars  seventy-five  cents  ;  what  did 
I  gain?  Ans.  $909.75. 

8.  Bought  a  barrel  of  flour  for  $  7.50,  three  bushels  of  rye 
for  $  2.75,  three  cords  of  wood  at  $  5.25  a  cord ;  I  sold  the 
flour  for  $  6.18,  the  rye  for  $  3.00,  and  the  wood  for  $  6.75  a 
cord  ;  what  was  gained  by  the  bargain  ?  Ans.  $  3.43. 

9.  A  young  lady  went  a  "  shopping."     Her  father  gave  her 
a  twenty-dollar  bill.     She  purchased  a  dress  for  $  8.16,  a  muff 
for  $  3.19,  a  pair  of  gloves  for  $  1.12,  a  pair  of  shoes  for  $  1.90, 
a  fan  for  $0.19,  and  a  bonnet  for  $  3.08  ;  how  much  money 
did  she  return  to  her  father  ?  Ans.  $  2.36. 

MULTIPLICATION. 
RULE. 

With  the  dollars,  cents,  (Sf-c.for  the  multiplicand,  proceed  as  in  Simpk 
Multiplication,  and  the  result  will  be  the  product  in  the  terms  of.  the 
lowest  denomination  contained  in  the  multiplicand.  If  the  multipli- 
cand consists  of  dollars  only,  the  product  will  be  dollars  ;  if  there  are 
cents,  either  with  or  without  dollars,  the  product  will  be  cents,  and  the 
two  right  hand  figures  must  be  separated  by  the  appropriate  point.  If 
there  are  mills,  the  product  will  be  mills,  and  tfie  three  right  hand  figures 
must  be  pointed  off.  The  figures  on  the  left  of  t/ie  point  will  denote 
dollars,  the  next  two  following  it  will  denote  centst  and  the  third  mills. 

EXAMPLES. 

1.  What  will  365  barrels  of  Genesee  flour  cost,  at  $  5.75  a 
barrel  ? 

$5.75 
365 

2875 
3450 
1725 

$2098.75  Ans. 


70  UNITED  STATES  MONEY.  [SECT.  xin. 

2.  What  will  128  pounds  of  sugar  cost,  at  13  cents  ~7  mills 
a  pound  ? 

$.13,7 
128 

•1096 
274 
137 

$  17.53,6  Ans. 

3.  What  will  126  pounds  of  butter  cost,  at  13  cents  a  pound  ? 

Ans.  $16.38. 

4.  What  will  63  pounds  of  tea  cost,  at  93  cents  a  pound  ? 

Ans.  jj  58.59. 

5.  What  will  43  tons  of  hay  cost,  at  13  dollars  75  cents  a 
ton?  Ans.  $591.25. 

6.  If  1  pound  of  pork  is  worth  7  cents  3  mills,  what  are  46 
pounds  worth  ?  Ans.  $  3.35,8. 

7.  If  Icwt.  of  beef  cost  3  dollars  28  cents,  what  are  76cwt. 
worth  ?  Ans.  $  249.28. 

8.  What  will  96,000  feet  of  boards  cost,  at  11  dollars  67  cents 
a  thousand  ?  Ans.  $  1120.32. 

9.  If  a  barrel  of  cider  be  sold  for  2  dollars  12  cents,  what 
will  be  the  value  of  169  barrels  ?  Ans.  $358.28. 

10.  What  will  be  the  value  of  a  hogshead  of  wine,  contain- 
ing 63gals.,  at  1  dollar  63  cents  a  gallon  ?       Ans.  $  102.69. 

11.  Sold  a  sack  of  hops,  weighing  396  pounds,  at  11  cents  3 
mills  a  pound  ;  to  what  did  it  amount  ?  Ans.  $  44.74,8. 

12.  Sold  19  cords  of  wood,  at  5  dollars  75  cents  a  cord  ;  to 
what  did  it  amount  ?  Ans.  $  109.25. 

13.  Sold  169  tons  of  timber,  at  4  dollars  68  cents  a  ton ; 
what  did  I  receive  ?  Ans.  $  790.92. 

14.  Sold  a  hogshead  of  sugar,  weighing  465  pounds ;  to  what 
did  it  amount,  at  14  cents  a  pound  ?  Ans.  $  65.10. 

15.  What  will  be  the  amount  of  789  pounds  of  leather,  at  18 
cents  a  pound  ?  Ans.  $  142.02. 

16.  What  will  be  the  expense  of  846  pounds  of  sheet  lead, 
at  5  cents  7  mills  a  pound  ?  Ans.  $  48.22,2. 

17.  When  potash  is  sold  for  132  dollars  55  cents  a  ton,  what 
will  be  the  price  of  369  tons  ?  Ans.  $  48910.95. 

18.  What  will  365  pounds  of  beeswax  cost,  at  18  cents  4 
mills  a  pound  ?  Ans.  $  67.16. 

19.  If  1  pound  of  tallow  cost  7  cents  3  mills,  what  are  968 
pounds  worth?  Ans.  $70.66,4. 


SECT,  mi.]  MULTIPLICATION.  71 

20.  What  will  a  chest  of  souchong  tea  be  worth,  containing 
69  pounds,  at  29  cents  9  mills  a  pound  ?         Ans.  820.63,1. 

21.  If  1  drum  of  figs  cost  2  dollars  75  cents,  what  will  be  the 
price  of  79  drums  ?  Ans.  $  217.25. 

22.  If  1  box  of  oranges  cost  6  dollars  71  cents,  what  will  be 
the  price  of  169  boxes  ?  Ans.  8  1133.99. 

23.  Purchased  796   pounds  of  cocoa,  at  11  cents  4  mills  a 
pound  ;  what  did  I  have  to  pay  ?  Ans.  8  90.74,4. 

24.  A  farmer  sold  691  bushels  of  wheat,  at  1  dollar  25  cents 
a  bushel ;  what  did  he  receive  for  it  ?  Ans.  8  863.75. 

25.  What  are  97  pounds  of  madder  worth,  at  17  cents  6  mills 
a  pound  ?  Ans.  $  17.07,2. 

26.  A  merchant  sold  73  hogsheads  of  molasses,  each  con- 
taining 63  gallons,  for  44  cents  a  gallon ;  how  much  money  did 
he  receive  ?  Ans.  8  2023.56. 

27.  A  drover  has  169  sheep,  which  he  values  at  2  dollars  69 
cents  a  head  ;  what  is  the  value  of  the  whole  drove  ? 

Ans.  8454.61. 

28.  A  farm  containing  144  acres  is  valued  at  69  dollars  74 
cents  8  mills  an  acre  ;  what  is  the  amount  of  the  whole  ? 

Ans.  $  10043.71,2. 

29.  An  auctioneer  sold  48  bags  of  cotton,  each  containing 
397  pounds,  at  13  cents  7  mills  a  pound  ;  what  is  the  value  of 
the  whole  ?  Ans.  $2610.67,2. 

30.  If  1  yard  of  broadcloth  cost  5  dollars  67  cents,  what  will 
be  the  value  of  48  yards  ?  Ans.  8  272.16. 

31.  A  wool-grower  has  179  sheep,  each  producing  4  pounds 
of  wool ;  what  will  be  its  value,  at  59  cents  3  mills  a  pound  ? 

Ans.  8424.58,8. 

32.  A  house  having  17  rooms  requires  6  rolls  of  paper  for 
each  room ;  now,  if  each  roll  cost  1  dollar  17  cents,  what  will 
be  the  expense  for  all  the  rooms  ?  Ans.  $  119.34. 

33.  What  will  89  yards  of  brown  sheeting  cost,  at  17  cents  a 
yard  ?  Ans.  815.13. 

34.  What  will  47,000  of  shingles  cost,  at  3  dollars  75  cents  a 
thousand  ?  Ans.  8  176.25. 

35.  Bought  47  hogsheads  of  salt,  each  containing  7  bushels, 
for  1  dollar  12  cents  a  bushel ;  what  did  it  cost  ? 

Ans.  8368.48. 

36.  What  will  a  ton  of  hay  cost,  at  1  dollar  17  cents  a  hun- 
dred weight  ?  Ans.  8  23.40. 

37.  If  1  foot  of  wood  cost  63  cents,  what  will  39  cords  cost  ? 

Ans.  $  196.56. 


72  UNITED  STATES  MONEY.  [SECT.  mi. 

38.  What  will  163  buckets  cost,  at  1  dollar  21  cents  a  bucket  ? 

Ans.  $  197.23. 

39.  What  will  78  barrels  of  shad  cost,  at  3  dollars  89  cents  a 
barrel?  Ans.  $303.42. 

40.  If  1  pound  of  salmon  cost  17  cents  5  mills,  what  will 
789  pounds  cost  ?  Ans.  $  138.07,5. 

41.  Bought  163  grindstones,  at  6  dollars  79  cents  each ;  what 
did  the  whole  cost  ?  Ans.  §  1106.77. 

42.  Sold  49  green  hides,  at  1  dollar  95  cents  each ;  what  did 
they  all  amount  to  ?  Ans.  895.55. 

43.  If  a  man's  wages  be  1  dollar  19  cents  a  day,  what  are 
they  for  a  year  ?  Ans.  $  434.35. 

44.  Hired  a  horse  and  chaise  to  go  a  journey  of  146  miles, 
at  16  cents  a  mile  ;  what  did  it  cost  ?  Ans.  $  23.36. 

45.  If  it  be  worth  3  dollars  68  cents  to  plough  one  acre  of 
land,  what  would  it  be  worth  to  plough  79  acres  ? 

Ans.  $290.72. 

46.  What  will  148  tons  of  plaster  of  Paris  cost,  at  2  dollars 
28  cents  a  ton  ?  Ans.  8  337.44. 

47.  Bought  79  tons  of  logwood,  at  49  dollars  75  cents  a  ton ; 
what  did  it  cost  ?  Ans.  8  3930.25. 

48.  Bought  5  gross  bottles  of  castor  oil,  at  37  cents  a  bottle  ; 
what  did  it  cost  ?  Ans.  $  266.40. 

49.  Sold  19  dozen  pair  of  men's  gloves,  at  47  cents  a  pair ; 
to  what  did  they  amount  ?  Ans.  8  107.16. 

50.  Bought  a  hogshead  of  wine  for  97  cents  a  gallon,  and 
sold  it  for  1  dollar  75  cents  a  gallon ;  what  did  I  gain  ? 

Ans.  849.14. 

51.  Bought  75  barrels  of  flour,  at  5  dollars  75  cents  a  barrel, 
and  sold  it  at  6  dollars  37  cents ;  what  did  I  gain  ? 

Ans.  $46.50. 

52.  Bought  17  score  of  penknives,  at  17  cents  each ;  what 
did  they  all  cost  ?  Ans.  $  57.80. 

53.  What  will  17£  tons  of  coal  cost,  at  $  9.62  per  ton  ? 

Ans.  8  168.35. 

54.  What  will  19  barrels  of  cider  cost,  at  8  1.37£  per  barrel  ? 

Ans.  826.12,5. 

DIVISION. 
RULE. 

With  the  sum  given  for  the  dividend,  proceed  as  in  Simple  Division, 
and  the  result  will  be  the  quotient  in  the  lowest  denomination  contained 
in  the  dividend. 


SECT,  xin.]  DIVISION.  73 

The  rule  for  pointing  off  cents  and  mills  is  the  same  as  in  Multiplication. 

If  the  dividend  consist  of  dollars  only,  and  be  either  smaller  than  tJie 
divisor,  or  not  divisible  by  it  without  -a  remainder,  annex  two  or  three 
ciphers,  as  the  case  may  require,  and  tJie  quotient  will  be  cents  or  mills 
accordingly. 

1.  If  97  bushels  of  wheat  cost  $  147.82,8,  what  is  the  value 
of  one  bushel  ?  Ans.  $  1.52,4. 

OPERATIpN. 

97)  147.82,8($  1.52,4 
97_ 
508 

485 

232 
194 

388 
388 

2.  Bought  1789  acres  of  land  for  $  1699.55  ;  what  cost  one 
acre?  Ans.  $0.95. 

3.  A  trader  sold  425  pounds  of  sugar  for  $  51.00  ;  what  was 
the  cost  of  one  pound  ?  Ans.  $  0. 12. 

4.  When  rye  is  sold  at  the  rate  of  628  bushels  for  $  471.00, 
what  is  that  a  bushel  ?  Ans.  $  0.75. 

5.  A  merchant  bought  329  yards  of  broadcloth  for  $  904.75  ; 
what  cost  one  yard  ?  Ans.  $  2.75. 

6.  When  a  chest  of  tea  containing  42  pounds  can  be  bought 
for  $  31.50,  what  cost  one  pound  ?  Ans.  $  0.75. 

7.  If  it  cost  $1460  to    support  a  family  365  days,   what 
would  be  the  expense  per  day  ?  Ans.  $  4.00. 

8.  A  shoe-dealer  sold  125  cases  of  shoes  for  $  2500  ;  what 
was  the  cost  per  case  ?  Ans.  $  20.00. 

9.  A  flour-merchant  sold  475  barrels  of  flour  for  $2018.75; 
what  cost  one  barrel  ?  Ans.  $  4.25. 

10.  Bought  42  barrels  of  pears  for  $  73.50  ;  what  cost  one 
barrel  ?  Ans.  $  1.75. 

11.  If  1624  pounds  of    pork  cost  $97.44,  what  cost  one 
pound  ?  Ans.  $  0.06. 

12.  If  47000  shingles  cost  $  176.25,  what  is  the  cost  per 
thousand  ?  Ans.  $  3.75. 

13.  Bought  148  tons  of  plaster  of  Paris  for  $  337.44  ;  what 
was  it  per  ton  ?  Ans.  $  2.28. 

14.  If  78  barrels  of  fish  cost  $  303.42,  what  will  one  barrel 
cost?  Ans.  $3.89. 

7 


74  UNITED  STATES  MONEY.  [SECT.  xin. 

15.  A  farmer  sold  691  bushels  of  wheat  for  $  863.75 ;  what 
was  it  per  bushel  ?  Ans.  $  1.25. 

16.  If  a  man  earn  $  434.35  in  a  year,  what  is  that  per  day  ? 

Ans.  $1.19. 

17.  Sold  169  tons  of  timber  for  $  790.92  ;  what  cost  one 
ton  ?  Ans.  $  4.68. 

18.  What  cost  one  pound  of  leather,  if  789  pounds   cost 
$142.02?  Ans.  $0.18. 

19.  If  369  tons  of  potash  cost  $48910.95,  what  will  be  the 
price  of  one  ton  ?  Ans.  $  132.55. 

20.  Bought  47  hogsheads  of  salt,  each  hogshead  containing 
7  bushels,  for  $  368.48  ;  what  cost  one  bushel  ?    Ans.  $  1.12. 

21.  If  19  cords  of  wood  cost  $  106.97,  what  cost  one  cord  ? 

Ans.  $5.63. 

22.  When  19  bushels  of  salt  can  be  bought  for  $  30.87,5, 
what  cost  one  bushel?  Ans.  $  1.62,5. 

23.  If  17  chests  of  souchong  tea,  each  weighing  59  pounds, 
cost  $ 672.01,  what  cost  one  pound?  Ans.  $ 0.67. 

24.  Sold  73  tons  of  timber  for  $  414.64 ;  what  did  I  receive 
per  ton  ?  Ans.  $  5.68. 

25.  Bought  oil  at  the  rate  of  144  gallons  for  $  234.00 ;  what 
did  I  give  per  gallon  ?  Ans.  $1.62,5. 

26.  A  landholder  sold  47  acres  of  land  for  $  1774.25 ;  what 
did  he  receive  per  acre  ?  Ans.  $  37.75 

27.  What  is  the  price  of  one  yard  of  broadcloth,  if  163  yards 
cost  $1106.77?  Ans.  $6.79. 

28.  If  a  farm,  containing  144  acres,  is  valued  at  $  10043.71,2, 
what  is  one  acre  worth  ?  Ans.  $  69.74,8. 

BILLS. 

1.  Boston,  July  4,  1835. 
Mr.  James  Dow, 

Bought  of  Dennis  Sharp, 

17  yds.  Flannel,  at  .45  cts. 

19    "    Shalloon,  "  .37    " 

16  "    Blue  Camlet,  "  .46    " 
13    "    Silk  Vesting,  "  .87    " 

9  "  Cambric  Muslin,  "  .63  " 

25  "  Bombazine,  "  .56  " 

17  "  Ticking,  "  .31  " 
19  "  Striped  Jean,  .16  " 

~~$61.33 
Received  payment, 

Dennis  Sharp. 


SECT.   XIII.] 

2. 

Mr.  Samuel  Smith, 

13  Ibs.  Tea, 
16    "   Coffee, 
Sugar, 


BILLS. 


75 


36 
47 
12 
7 
13 


"  Cheese, 
"  Pepper, 
"  Ginger, 


Haverhill,  May  5,  1835. 


Bought  of  David  Johnson, 
.98  cts. 
.15  " 
.13   " 
.09   " 
.19   " 


"    Chocolate, 
Received  payment, 


at 

u 

it 
u 

tl 
u 
II 


.17   " 
.61    " 


$35.45. 


David  Johnson. 


3. 

Mr.  John  Dow, 


17  yds.  Broadcloth, 


Salem,  February  29,  1835. 
Bought  of  Richard  Fuller, 


29 
60 
49 
18 
27 
75 
36 
49 


Cassimere, 
Bleached  Shirting, 
Ticking, 
Blue  Cloth,  . 
Habit  do. 
Flannel, 
Plaid  Prints, 
Brown  Sheeting, 


Received  payment, 


$5.25 
1.62 
.17 
.27 
3.19 
2.75 
.61 
.75 
.18 

"8372.90. 
Richard  Fuller. 


4. 

Mr.  John  Rilley, 


10  pair  Boots, 
19     "    Shoes, 
83     «    Hose, 
47  Ibs.  Ginger, 
91    "     Chocolate, 
47   "     Pepper, 
68   "     Flour, 
27  pair  Gloves, 


Received  payment, 


Baltimore,  January  20,  1835. 

Bought  of  James  Somes, 
at         $2.75 
"  1.25 

"  1.29 

"  .17 

"  .39 

"  .23 

.13 
"  1.39 

~$  258.98. 


James  Somes. 


UNITED  STATES  MONEY.  [SECT.  xin. 

5.  Philadelphia,  June  11,  1835. 
Mr.  Moses  Thomas, 

Bought  of  Luke  Dow, 

27  National  Spelling-Books,         at     $  0.  19 
25  Parker's  Composition,  .27 

17  National  Arithmetics,  .75 
9  Greek  Lexicons,                                 3.75 
8  Ainsworth's  Dictionaries,         "         4.50 

27  Greek  Readers,  "  2.25 

18  Folio  Bibles,  "  9.87 
75  Leverett's  Caesar,  "  .31 
67  Fisk's  Greek  Grammar,  .75 
15  Folsom's  Cicero's  Orations,  "  1.12 

$423.09. 
Received  payment, 

Luke  Dow,  by 

Timothy  True. 

6.  Boston,  June  26,  1835. 
Dr.  Enoch  Cross, 

Bought  of  Maynard  &  Noyes, 

14  oz.  Ipecacuanha,  at  $  0.67 

23  "    Laudanum,  "  .89 

17  "    Emetic  Tartar,  "  1.25 

25  "    Cantharides,  "  2.17 

27  «    Gum  Mastic,  "  .61 
56  "    Gum  Camphor, 

8  136.94. 
Received  payment, 

Maynard  &  Noyes, 

by  Timothy  Jones. 

7.  Newburyport,  June  5,  1835. 
Mr.  John  Somes, 

Bought  of  Samuel  Gridley, 

7£  yds.  Broadcloth,  at  $4.50 

16flbs.    Coffee,    -  "  .16 

18J  "      Candles,  "  .25 

30     "      Soap,  "  .17 

3     "      Pepper,  "  .19 

Ginger,  "  .18 


" 


Received  payment, 

Samuel  Gridley. 


SECT.  XIII.] 


BILLS. 


77 


8. 
Mr.  Benjamin  Treat, 


37  Chests  Green  Tea, 

41      "      Black    do. 

40      "      Chests  of  Imperial  Tea, 

13  Crates  Liverpool  Ware, 

Received  payment, 


Boston,  May  1,  1835. 

Bought  of  John  True, 
at     $  25.50 
16.17 
97.75 
169.37 


$7718.28. 
John  True. 


9. 

Mr.  John  Cummings, 


97  bbl.  Genesee  Flour, 

167    "   Philadelphia  do. 

87    "    Baltimore      do. 

196  "   Richmond     do. 
275   «    Howard  St  do. 

69  bu.  Rye, 
136  "    Virginia  Corn, 

68  "    North  River  do. 
169  «    Wheat, 

76TonLehighCoal, 

89    "    Iron, 

49  Grindstones, 

39  Pitchforks, 

197  Rakes, 
86  Hoes, 
78  Shovels, 

187  Spades, 
91  Ploughs, 
83  Harrows, 
47  Handsaws, 
35  Millsaws, 
47  cwt.  Steel, 
57    "   Lead, 

Received  payment, 


New  York,  July  11,  1835. 
Bought  of  Lord  &  Secomb. 


at 


$6.25 
5.95 
6.07 
5.75 
7.25 
1.16 
.67 
.76 
1.37 
9.67 
69.70 
3.47 
1.61 
.17 
.69 
1.17 
.85 
11.61 
17.15 
3.16 
18.15 
9.47 
6.83 


17315.32. 


Lord  &  Secomb. 


78  COMPOUND  MULTIPLICATION.  [SECT.  xiv. 

SECTION  XIV. 
COMPOUND   MULTIPLICATION. 

CASE  I. 

COMPOUND  MULTIPLICATION  consists  in  multiplying  numbers 
of  different  denominations  by  simple  numbers. 

I.  What  will  6  bales  of  cloth  cost,  at  7«£.  12s.  7d.  per  bale  ? 
£       8      d         In  this  question,  we  multiply  7d.  by  6,  and  find 
7     12     7     ^e  product  to  be  42d.     This  we  divide  by  12,  the 

g     number  of  pence  in  a  shilling,  and  find  it  contains 
JJT — TjT—       3s.  and  6d.     We  write  the  6d.  under  the  pence, 
and  carry  3  to  the  product  of  6  times  12,  and  find 
the  amount  to  be  75s.,  which  we  reduce  to  pounds  by  dividing 
them  by  20,  and  find  them  to  be  3£.  15s.     We  write  down  the 
shillings  under  the  shillings,  and  carry  3  to  the  product  of  6 
times  7<£. ;  and  we  thus  find  the  answer  to  be  45£.  15s.  6d. 
From  the  above  illustration  we  deduce  the  following 

RULE. 

When  the  multiplier  is  less  than  12,  multiply  by  the  multiplier  and 
carry  as  in  Compound  Addition. 

NOTE.— For  the  answers  in  Multiplication,  see  Section  XV.,  in  Di- 
vision. 

2.  What  cost  9yds.  of  cloth,  at  1  £.  3s.  8d.  per  yard  ? 

Ans.   10£.  13s.  Od. 

3.  What  cost  7bbls.  of  flour,  at  !<£.  8s.  7£d.  per  barrel  ? 

4.  What  cost  81bs.  of  Cayenne  pepper,  at  7s.  9£d.  per  Ib.  ? 

5.  Multiply  10yd.  3qr.  3na.  by  5. 

6.  Multiply  3cwt.  Iqr.  81b.  by  9. 

7.  Multiply  7T.  llcwt.  Iqr.  201b.  by  5. 

8.  Multiply  7  days  15h.  35m.  18sec.  by  10. 

9.  Multiply  18<£.  16s.  74-d.  by  4.  Ans.  75<£.  6s.  6d. 
10.  Multiply  15=£.  11s.  8|d.  by  8.     Ans.  124c£.  13s.  lOd. 

II.  Multiply  27£.  19s.  ll£d.  by  9.     Ans.  251,£.  19s.  7£d. 

12.  Multiply  19<£.  5s.  7£d.  by  11.        Ans.  212£.  Is.  7|d. 

13.  Multiply  81£.  14s.  9d.  by  8.         Ans.  653^.  18s.  Od. 

14.  Multiply  15£.  18s.  5d.  by  7.         Ans.  111£.  8s.  lid. 

15.  Multiply  13=£.  5s.  4|d.  by  12.          Ans.  159^.  4s.  9d. 

16.  Multiply  171b.  7oz.  13dwt.  13gr.  by  9. 

17.  Multiply  151b.  lloz.  19dwt.  15gr.  by  7. 


SECT,  xiv.]  COMPOUND  MULTIPLICATION.  79 

18.  Multiply  16T.  12cwt.  3qr.  131b.  12oz.  by  11. 

19.  Multiply  13T.  3cwt.  Iqr.  141b.  13oz.  by  8. 

20.  Multiply  21b  5i  53  19   16£gr.  by  8. 

21.  Multiply  47yd.  3qr.  2na.  2in.  by  7. 

22.  Multiply  17m.  7fur.  36rd.  13ft.  7in.  by  12. 

23.  Multiply  16deg.  39m.  3fur.  39rd.  5yd.  2ft.  by  9. 

24.  Multiply  16deg.  20m.  7fur.  12rd.  8ft.  llin.  IJ-bar.  by  6. 

25.  Multiply  16A.  2R.  4p.  19yd.  7ft.  79in.  by  11. 

26.  Multiply  7  cords  1  16ft.  1629m.  by  4. 

27.  Multiply  29hhd.  61gal.  3qt.  Ipt.  3gi.  by  7. 

28.  Multiply  3  tuns  3hhd.  56gal.  2qt.  by  9. 

29.  Multiply  7hhd.  5gal.  2qt.  Ipt.  by  8. 

30.  Multiply  19bu.  2pk.  7qt.  Ipt.  by  6. 

31.  Multiply  36ch.  18bu.  3pk.  7qt.  by  7. 

32.  Multiply  13y.  316d.  15h.  27m.  39sec.  by  8. 

33.  If  a  man  gives  each  of  his  9  sons  23A.  3R.  19£p.,  what 
do  they  all  receive  ? 

34.  If  12  men  perform  a  piece  of  labor  in  7h.  24m.  30sec., 
how  long  would  it  take  1  man  to  perform  the  same  task  ? 

35.  If  1  bag  contain  3bu.  2pk.  4qt.,  what  quantity  do  8  bags 
contain  ? 

CASE  II. 

When  the  multiplier  is  more  than  12,  and  is  a  composite  num- 
ber, that  is,  a  number  which  is  the  product  of  two  or  more 
numbers,  the  question  is  performed  as  in  the  following 

EXAMPLE. 

36.  What  will  42  yards  of  cloth  cost,  at  6s.  9d.  a  yard  ? 

£.  e.  d.  In  this  example,  we  find  that  6 

069  multiplied  by  7  will  produce  the 

6  quantity  42  yards.  We  therefore 

206  =  price  of  6  yds.  multiply  6s  9d.  first  by  the  6,  and 

•y  then  its  product  by  7  ;  and  the  last 

uct,  14£.  3s.  6d.  is  the  answer 


A     o 
14     3     6  = 

The  pupil  will  now  see  the  propriety  of  the  following 

• 
RULE. 

Multiply  by  one  of  the  factors  of  the  composite  number,  and  tlte  prod- 
uct thus  obtained  by  the  other. 

37.  What  will  16  yards  of  velvet  cost,  at  3s.  8d.  per  yard  ? 


80  COMPOUND  MULTIPLICATION.  [SECT.  sir. 

38.  What  will  72  yards  of  broadcloth  cost,  at  19s.  lid.  per 
yard  ? 

39.  What  will  84  yards  of  cotton  cost,  at  Is.  lid.  per  yard  ? 

40.  Bought  90  hogsheads  of  sugar,  each  weighing  12cwt. 
2qr.  1  lib.  ;  what  was  the  weight  of  the  whole  ? 

41.  What  cost  18  sheep,  at  5s.  9£d.  a  piece  ? 

42.  What  cost  21  yards  of  cloth,  at  9s.  lid.  per  yard  ? 

43.  What  cost  22  hats,  at  11s.  6d.  each  ? 

44.  If  1  share  in  a  certain  stock  be  valued  at  13£.  8s.  9Jd., 
what  is  the  value  of  96  shares  ? 

45.  If  1  spoon  weigh  3oz.  5dwt  15gr.,  what  is  the  weight  of 
120  spoons  ? 

46.  If  a  man  travel  24m.  7fur.  4rd.  in  1  day,  how  far  will  he 
go  in  1  month  ? 

47.  If  the  earth  revolve  0°  15'  per  minute,  how  far  per  hour  ? 

48.  Multiply  39A.  3R.  17p.  30yd.  8ft.  lOOin.  by  32. 

49.  If  a  man  be  2d.  5h.  17m.  19sec.  in  walking  1  degree, 
how  long  would  it  take  him  to  walk  round  the  earth,  allowing 

days  to  a  year  ? 


CASE  III. 

When  the  multiplier  is  such  a  number  as  cannot  be  pro- 
duced by  the  product  of  two  or  more  numbers,  we  should  pro- 
ceed as  in  the  following 

EXAMPLE. 

50.  What  is  the  value  of  53  tons  of  iron,  at  I&£.  17s.  lid.  a 
ton? 

£.        a.      d.  £.      s.     d. 

18     17  11  18  17  11 

_  5  __  3 

94      9    7  =  price  of  5  tons.  56  13    9  =  price  of  3  tons. 
_  10 

944     15  10=  price  of  50  tons.  Because   53    is   a    prime 

56     13     9  =  price  of  3  tons,  number,  that  is,  it  cannot  be 

1001       9     7  =  price  of  53  tons.  Prodfuced  b\the  Produ^  of 

any  two  numbers  ;  we  there- 

fore find  a  convenient  composite  number  less  than  the  given 
number,  viz.  50,  which  may  be  produced  by  multiplying  5  by 
10.  Having  found  the  price  of  50  tons  by  the  last  Case,  we 
then  find  the  price  of  the  3  remaining  tons  by  Case  I.,  and 
add  it  to  the  former,  making  the  value  of  the  whole  quantity 
9s.  7d. 


SECT,  xiv.]  BILLS.  81 

The  pupil  will  hence  perceive  the  propriety  of  the  following 

RULE. 

Take  for  successive  multipliers  two  or  more  numbers,  whose  continued 
product  will  be  nearest  the  proper  multiplier,  and  then  find  the  value  of 
the  remainder  by  Case  I.,  and  the  sum  of  the  last  two  products  ivitt  be 
the  answer. 

51.  What  will  57  gallons  of  wine  cost,  at  8s.  3£d.  per  gallon  ? 

52.  Bought  29  lots  of  wild  land,  each  containing  117A.  3R. 
27p.  ;  what  were  the  contents  of  the  whole  ? 

53.  Bought  89  pieces  of  cloth,  each  containing  37yd.  3qr. 
2na.  2in.  ;  what  was  the  whole  quantity  ? 

54.  Bought  59  casks  of  wine,  each  containing  47gal.  3qt.  Ipt.  ; 
what  was  the  whole  quantity  ? 

55.  If  a  man  travel  17m.  3fur.  13rd.  14ft.  in  one  day,  how 
far  will  he  travel  in  a  year  ? 

56.  If  a  man  drink  3gal.  Iqt.  Ipt.  of  beer  in  a  week,  how 
much  will  he  drink  in  52  weeks  ? 

57.  There  are   17  sticks  of  timber,  each    containing  37ft. 
978in.  ;  what  is  the  whole  quantity  ? 

58.  There    are  17  piles  of  wood,  each  containing  7  cords 
98  cubic  feet  ;  what  is  the  whole  quantity  ? 

59.  Multiply  2hhd.  19gal.  Oqt.  Ipt.  by  39. 

60.  Multiply  3bu.  Ipk.  4qt.  Ipt.  Igi.  by  53. 

61.  Multiply  16ch.  7bu.  2pk.  Oqt.  Opt.  by  17. 

BILLS. 

1.  London,  July  4,  1835. 
Dow,  Vance,  &  Co.,  of  Boston,  U.  S., 

Bought  of  Samuel  Snow, 

45  yds.  Broadcloth,  at        8s.     4d. 

50    "           "  "  10s.     6d. 

56    "           "  "         3s.     7£d. 

63    "           "  "  12s.  llfd. 

72    «          «  «  19s.  lid. 

81    "          "  "         9s.     3d. 

35    "           "  "  19s.     7£d. 

99    "          "  "  16s.     0£d. 

66    "          "  "         8s.  lid. 

33    "          «  "  16s. 


~376<£.  7s.  Of  d. 
Received  payment, 

Samuel  Snow. 


82  COMPOUND  MULTIPLICATION.  [SECT.  xiv. 

2.  Quebec,  Jan.  8,  1835. 

Mr.  John  Vose,  Bought  of  Vans  &  Conant, 

46  Ivory  Combs,  at         3s.     5£d. 

47  Ibs.  Colored  Thread,  "       .  6s.     9jd. 

51  yds.  Durant,  "         Is.     8d. 

52  Silk  Vests,  "         6s.     7d. 

53  Leghorns,  "       11s.     9^d. 

57  ps.  Nankin,  "         8s. 

58  Ibs.  White  Thread,    "         9s. 


~~128£.   16s. 
Received  payment,       Vans  &  Conant. 

3.  Montreal,  July  4,  1835. 
Mr.  James  Savage,                 Bought  of  Joseph  Dowe, 

83  gals.  Lisbon  Wine,      at  6s.  7d. 

85  "     Port         do.          "  3s.  9£d. 

86  "      Madeira  do.          "  4s.  ll£d. 

87  "     Temperance  do.  "  3s.  6£d. 
89    "     Oil,                       "  5s.  3d. 

91  Leghorns,  "     19s.  10£d. 

92  Ibs.    Green  Tea,          "       3s.     l£d. 
$3  pair  Thread  Hose,       "       4s.     4£d. 

94  "      Silk  Gloves,         "       3s.     3£d. 

95  "      Silk  Hose,  "       6s.     6£d. 

97  yds.    Linen,  "       5s.     5Jd. 

98  gals.  Winter  Strained  Oil,    7s.     7£d. 

^33a£.  19s. 
Received  payment,  Joseph  Dowe. 

4.  Montreal,  June  17,  1835. 
Mr.  Samuel  Simpson,     Bought  of  Lackington,  Grey,  &  Co. 

19yds.  Cloth,  at  Is.  6d. 

23  "  Worsted,  "  7s.  8£d. 

26  "  Baize,  «  3s.  ll^d. 

29  «  Camlet,  «  6s.  10|d. 

31  "  Bombazine,  "  Is.  5£d. 

34  "  Linen,  "  3s.  7d. 

37  "  Cotton,  "  11s.  9d. 

38  "  Flannel,  "  6s.  lid. 

39  "  Calico,  "  3s.  lO^d. 
41  "  Broadcloth,  "  6s.  9^d. 
43  "  Nankin,  "  7s.  5fd. 

~106c£.  Is. 
Received  payment,        Lackington,  Grey,  &  Co. 


SECT,  xiv.]  BILLS.  83 

5.  Liverpool,  June  2,  1835. 

John  Jones,  of  Philadelphia,  U.  S., 

Bought  of  Thomas  Hasseltine, 
297  yds.  Black  Broadcloth,     at     17s.     3£d. 
473    "    Blue          do.  "       9s.  ll£d. 

512    "    Red  do.  "     15s.  lOd. 

624    "    Green        do.  "     12s.     8d. 

765    "    White        do.  "     19s.    9|d. 

169    "    Black  Velvet,  tc     13s.     5£d. 

698    "    Green      do.  "     15s.     6fd. 

315    "    Red          do.  "     14s.     3£d. 

713    "    White      do.  "     11s.     7£d. 

519    "    Carpet,  "     13s.     6£d. 

147    "    Black  Kerseymere,  "     16s.    7fd. 
386    "    Blue  do.          "     14s.     3£d. 

137    "    Green          do.          "     19s.     9d. 
999    "    Black  Silk,  "     15s.     8d. 

~~5012<£.  Os.  lljd. 
Received  payment, 

Thomas  Hasseltine. 


6.  London,  May  11,  1846. 

Messrs.  Kimball,  Jewett,  &  Co.,  of  Boston,  U.  S., 

Bought  of  Benjamin  Fowler, 

£.      s.  d. 

2345  yds.  Red  Broadcloth,  at     l£.  17s.     9£d.  =    4428     12  7£ 

7186    "     Green      do.          "     3£.  15s.     8£d.  =  27202       0  1 

8011    "     Black      do.          "     2£.  18s.  10£d.  =  23574       0  8| 

6789    "     Blue        do.          "     l£.     6s.     9d.    =    9080       5  9 

3178    "     White    do.          "     2£.     Is.     7fd.  =     6617     10  5^ 

2365    "     Pongee  Silk,        ««     l£.     2s.     S^d.  =     2685      5  2^ 

5107    "     Black      do.          4<     l£.     7s.     5|d.  =    7006       3  3| 

4444  bales  Cotton  Cloth,       «     3£.  16s.     8^d.  =  17039     19  3 

7777     <c    Irish  Linen,          "  17£.   19s.     9d.    ==139888     15  9 

1234     "    Nankin,  ««     7£.  15s.   lid.    =    9620       1  2 

4567     "    Flannel,  "     S£.   16s.     7^d.  =  40332       6  4£ 

9876     "    Bombazine,  {<     3£.     5s.     5|d.  =  32333     12  3 

7658     "    Calico,  "     9£.  17s.     6£d.  =  75638     14  1 

8107     "    Camlet,  ««     8£.     6s.     OJd.  =  67313       8  8| 

4725     {<     Baize,  «     3£.     7s.     9|d.  =  16020     14  0| 

5670     "    Durant,  "     4£.  18s.     5|d.  =  27907      0  U 

506688  .£.  10s.  4  |d. 
Received  payment, 

Benjamin  Fowler. 


84  COMPOUND  DIVISION  [SECT.  xv. 

SECTION  XV. 
COMPOUND  DIVISION. 

COMPOUND  DIVISION  is  when  the  dividend  consists  of  several 
denominations. 

EXAMPLES. 

1.  Divide  59&£.  8s.  9d.  equally  among  5  persons. 

£  B-  d.  Having  divided  the  pounds  by  5,  we  find 
5)598  8  9  3f .  remaining,  which  are  60s. ;  to  these  we 

jjg     jg — <j     add  the  8s.  in  the  question,  and  again  divide 

by  5  and  find  3s.  remaining,  which  are  36d.  ; 

to  these  we  add  the    9d.  in  the  question,  and  divide  their  sum 

by  5.     The  several  quotients  we  write  under  their  respective 

denominations. 

2.  Divide  168£.  15s.  Od.  equally  among  36  men. 

o£\v£o     I°K.     n/ ,14?        When  the  divisor  is  more  than  12,  we 

4*'  usually   perform  the  operation  by  Long 

Division.     In  the  present  example,   we 

24  first  divide  the  pounds  by  36,  and  obtain 

4£.  for  the  quotient  and  24<£.  remaining, 

36)495(13s.  which  we  reduce  to  shillings  and  annex 

36  the  15s.,  and  again  divide  by  36,  and  ob- 

"J35  tain  13s.  for  the  quotient.    The  remainder 

10Q  we  reduce  to  pence,  and  again  divide,  and 

~2=  obtain  9d.  for  the  quotient. 

12 

36)324(9d. 
324 

From  the  above  examples  and  illustrations  we  deduce  the 
following 

RULE. 

Divide  tlte  highest  denomination  of  the  dividend  by  the  divisor,  and, 
if  there  be  a  remainder,  reduce  it  to  the  next  lower  denomination,  adding 
to  the  number  thus  found  the  number  in  the  dividend  of  the  same  de- 
nomination. Divide  the  result  thus  obtained  by  the  divisor ;  and,  if 
there  be  a  remainder,  proceed  as  before,  till  all  the  denominations  of  the 
dividend  are  taken,  or  till  the  work  is  finished.  The  successive  quo- 
tients will  be  of  the  same  denominations  with,  the  successive  numbers 
divided,  or  will  correspond  with  the  several  denominations  of  the  divi- 
dend. 


SECT,  xv.]  COMPOUND  DIVISION.  85 

NOTE.  —  When  the  operation  is  performed    by   Short  Division,  the 
several  quotients  must  be  placed  under  their  respective  denominations. 

2.  When  10<£ .  13s.  Od.  are  paid  for  9  barrels  of  flour,  what 
is  the  cost  of  one  barrel  ? 

3.  If  7bbls.  of  flour  cost  10,£.  Os.  4£d.,  what  cost  one  barrel  ? 

4.  Paid  3<£.  2s.  4d.  for  81bs.  of  Cayenne  pepper ;  what  was 
the  cost  of  one  pound  ? 

5.  Divide  54yd.  2qr.  3na.  equally  among  5  persons. 

6.  Divide  29cwt.  3qr.  161b.  equally  among  9  persons. 

7.  If  5  pair  of  oxen  in  one  year  consume  37  tons  17cwt.  Oqr. 
16  Ib.  of  hay,  what  quantity  would  be  sufficient  for  one  pair  ? 

8.  If  one  man  could  perform  a  piece  of  labor  in  76  days  llh. 
53m.,  how  long  would  it   take  10  men  to  perform  the  same 
labor  ? 

9.  Divide  15L£.  19s.  ll£d.  by  9.      Ans.  16,£.  17s.  9Jd. 

10.  Divide  350£.  17s.  3£d.  by  519.  Ans.  13s.  6£d. 

11.  Divide  225<£.  Is.  10£d.  by  63.        Ans.  3<£.  11s.  5£d. 

12.  Divide  159<£.  4s.  9d.  by  12.  Ans.  13£.  5s.  4|d. 

13.  Divide  75o£.  6s.  6d.  by  4.  Ans.   18£.  16s.  7£d. 

14.  Divide  lll£.  8s.  lid.  by  7. 

15.  Divide  159£.  4s.  9d.  by  12. 

16.  Divide  1581b.  9oz.  Idwt.  21gr.  by  9. 

17.  Divide  11  lib.  lloz.  17dwt.  9gr.  by  7. 

18.  Divide  183  tons  Icwt.  2qr.  lllb.  4oz.  by  11. 

19.  Divide  105  tons  7cwt.  Oqr.  61b.  8oz.  by  8. 

20.  Divide  191b  9 i  43  29  9gr.  by  8. 

21.  Divide  335yd.  2qr.  Ona.  0£in.  by  7. 

22.  Divide  215m.  7fur.  Ird.  14ft.  6in.  by  12. 

23.  Divide  149deg.  8m.  Ofur.  Ord.  lyd.  1ft.  6in.  by  9. 

24.  Divide  97deg.  55m.  7fur.  35rd.  4ft.  2in.  Ibar.  by  6. 

25.  Divide  181A.  3R.  lip.  6yd.  4ft.  41in.  by  11. 

26.  Divide  31  cords  83ft.  1332in.  by  4. 

27.  Divide  209hhd.  55gal.  3qt.  Opt.  Igi.  by  7. 

28.  Divide  35  tuns  3hhd.  4gat.  2qt.  by  9. 

29.  Divide  56hhd.  45gal.  by  8. 

30.  Divide  118bu.  Ipk.  5qt.  by  6. 

31.  Divide  255ch.  24bu.  3pk.  Iqt.  by  7. 

32.  Divide  HOy.  343d.  3h.  41m.  12sec.  by  8. 

33.  A  man  divides  his  farm  of  214A.  3R.   12p.  equally 
among  his  9  sons ;  how  much  does  each  receive  ? 

34.  If  one  man  perform  a  certain  piece  of  labor  in  3da.  16h. 
54m.,  how  long  would  it  take  12  men  to  perform  the  same 
work  ? 

8 


86  COMPOUND  DIVISION.  [SECT.  TV. 

35.  A  farmer  has  29  bushels  of  rye,  which  he  wishes  to  put 
into  8  bags ;  how  much  must  each  bag  contain  ? 

CASE  II. 

When  the  divisor  is  a  composite  number,  proceed  as  in  the 
following 

EXAMPLE. 

36.  If  42  yards  of  cloth  cost  14£.  3s.  6d.,  what  is  the  value 
of  1  yard  ?    "  Ans.  6s.  9d. 

£.      B.     d.         In  this  question,  we  find  the  component  parts 
7)14     3     6     of  42  are  6  and  7  ;  we  therefore  first  divide  the 
gTg     Q     Q     price  by  7,  and  then  divide  the  quotient  by  6. 

069 

From  the  above,  we  deduce  the  following 

RULE. 

Divide  the  dividend  by  one  of  tlie  component  parts,  and  the  quotient 
thence  arising  by  the  other ,  and  the  last  quotient  will  be  the  answer. 

NOTE.  —  To  find  the  true  remainder;  multiply  the  last  remainder  by 
the  first  divisor,  and  to  the  product  add  the  first  remainder. 

37.  If  16  yards  of  velvet  cost  2£.  18s.  8d.,  what  will  1  yard 
cost  ? 

38.  If  72  yards  of  broadcloth  cost  71  £.  14s.  Od.,  what  is  the 
value  of  1  yard  ? 

39.  If  84  yards  of  cotton  cost  8£.  Is.  Od.,  what  will  1  yard 
cost  ? 

40.  If  90  hogsheads  of  sugar  weigh  56T.  13cwt.  3qr.  lOlb., 
what  is  the  weight  of  1  hogshead  ? 

41.  What  will  be  the  price  of  1  sheep,  if  18  cost  5<£.  4s.  3d.  ? 

42.  If  21  yards  of  cloth  cost  10<£.  8s.  3d.,  what  is  the  price 
of  1  yard  ? 

43.  What  is  the  value  of  1  hat,  when  22  cost  12£.  13s.  Od.  ? 

44.  When  96  shares  of  a  certain  stock  are  valued  at  1290<£. 
4s.  Od.,  what  would  be  the  cost  of  1  share  ? 

45.  If  120  spoons  weigh  321b.  9oz.   15dwt.,  what  does  1 
weigh  ? 

46.  If  a  man  in  1  month  travel  746m.  5fur.  Ord.,  how  far  does 
he  go  in  1  day  ? 

47.  If  the  earth  revolve  15°  on  its  axis  in  1  hour,  how  far 
does  it  revolve  in  1  minute  ? 


SECT,  xv.]  COMPOUND  DIVISION.  87 

48.  Divide  1275A.  2R.   16p.  22yd.  8ft.  32in.  equally  among 
32  men. 

49.  If  a  man  walk  round  the  earth  in  2y.  68d.   19h.  54m., 
how  long  would  it  take  him  to  walk  1  degree,  allowing  365£ 
days  to  a  year  ? 

The  following  questions  are  to  be  performed  as  the  second 
example  of  this  section. 

50.  If  53  tons  of  iron  cost  100  L£.  9s.  7d.,  what  is  the  value 
of  1  ton  ? 

51.  If  57  gallons  of  wine  cost  23<£.  lls.  5Jd.,  what  cost  1 
gallon  ? 

52.  Divide  3419A.  2R.  23p.  by  29. 

53.  If  89  pieces  of  cloth  contain  3375yds.  3qr.  Ina.  OJin., 
how  much  does  1  piece  contain  ? 

54.  If  59  casks  contain  44hhd.  52gal.  2qt.  Ipt.  of  wine,  what 
are  the  contents  of  1  cask  ? 

55.  If  a  man  travel  in  1  year  (365  days)  6357m.  V5fur.  14rd. 
ll^ft.,  how  far  is  that  per  day  ? 

56.  When  175gal.  2qt.  of  beer  are  drunk  in  52  weeks,  how 
much  is  consumed  in  1  week  ? 

57.  When  17  sticks  of  timber  measure  15T.  38ft.  1074in., 
how  many  feet  does  1  contain  ? 

58.  Divide  132  cords  2ft.  by  17. 

59.  Divide  89hhd.  52gal.  3qt.  Ipt.  by  39. 

60.  Divide  179bu.  3pk.  5qt.  Opt.  Igi.  by  53. 

61.  Divide  275ch.  19bu.  2pk.  equally  among  17  men. 

62.  Divide  796<£.  19s.  8d.  by  386.      Ans.  2<£.  Is.  3f ff d. 

63.  Divide  61S£.  16s.  7d.  by  571.      Ans.  l£.  Is.  8-^d. 

64.  Divide  1678£.  14s.  3d.  by  97.      Ans.  17£.  6s.  Iffd. 

65.  Divide  697T.  18cwt.  3qr.  141b.  by  146. 

Ans.  4T.  15cwt.  2qr.  12T2¥yb. 

66.  Divide  916m.  3fur.  30rd.  10ft.  6in.  by  47. 

Ans.  19m.  3fur.  39rd.  13ft.  2f  f  in. 

67.  Divide  718A.  3R.  37p.  by  29.      Ans.  24A.  3R.  6f|p. 

68.  Divide  815A.  1R.  17p.  200ft.  by  87. 

Ans.  9A.  1R.  19p.  139|fft. 

69.  Divide  144A.  3R.  18p.  3yd.  1ft.  36in.  by  11. 

Ans.  13A.  OR.  27p.  3yd.  Oft.  45T9Tin. 

70.  Divide  6718<£.  19s.  lid.  by  47. 

Ans.  U2£.  19s.  Iffd. 

71.  Divide  1237c£.  17s.  4d.  by  86.     Ans.  14£.  7s.  lOffd. 

72.  Purchased  1ST.  17cwt.  3qr.  201b.  of  copperas,  at  4  cents 
per  pound.     I  sold  4T.  6cwt.  Iqr.  141b.  at  5  cents  per  pound, 


88  COMPOUND   DIVISION.  [SECT.  xv. 

and  7T.  Icwt.  3qr.  lOlb.  at  6  cents  per  pound.  Moses  Atwood 
purchased  one  fourth  of  the  remainder  at  6  cents  per  pound. 
One  half  of  what  then  remained  I  sold  to  J.  Gale  at  10  cents 
per  pound.  The  remaining  quantity  I  sold  to  J.  Smith  at  12 
cents  per  pound ;  but  he  has  become  a  bankrupt,  and  I  lose 
half  my  debt.  What  have  I  gained  by  my  purchase  ? 

Ans.  $1001.34. 

QUESTIONS  TO  BE  PERFORMED  BY  ANALYSIS. 

1.  If  7  pair  of  shoes  cost  $  8.75,  what  will  one  pair  cost  ? 
what  will  20  pairs  cost  ?  Ans.  $  25.00. 

2.  If  5  tons  of  hay  cost  $  85,  what  will  1  ton  cost  ?  what  will 
17  tons  cost  ?  Ans.  $  289.00. 

3.  When  $  0.75  are  paid  for  3gal.  of  molasses,  what  is  the 
value  of  Igal.  ?     WThat  cost  37gal.  ?  Ans.  $  9.25. 

4.  Gave  $  1.92  for  41bs.  of  tea;  what  cost  lib.  ?  what  cost 
371bs.  ?  Ans.  $  17.76. 

5.  For  121bs.  of  rice  I  paid  $  1.08  ;  what  was  paid  for  lib. ; 
and  what  must  I  give  for  251bs.  ?  Ans.  $  2.25. 

6.  Gave  S.  Smith  $  63.00  for  9  tubs  of  butter ;  what  was  the 
cost  of  1  tub  ?     What  cost  27  tubs  ?  Ans.  8  189.00. 

7.  T.  Swan  can  walk  20  miles  in  5  hours ;  how  far  can  he 
walk  in  1  hour  ?     How  long  would  it  take  him  to  walk  from 
Bradford  to  Boston,  the  distance  being  in  a  straight  line  28 
miles  ?  Ans.  7  hours. 

8.  If  a  hungry  boy  would  eat  49  crackers  in  1  week,  how 
many  would  he  eat  in  1  day  ?     how  many  would  be  sufficient 
to  last  him  19  days  ?  Ans.  133  crackers. 

9.  Gave  $  20  for  5  barrels  of  flour ;  what  cost   1   barrel  ? 
what  cost  40  barrels  ?  Ans.  $  160.00. 

10.  For  31bs.  of  lard  there  were  paid  36  cents  ;  what  was  the 
cost  of  37ibs.  ?  Ans.. $4.44. 

11.  Paid  F.  Johnson  72  cents  for  9  nutmegs ;  how  many  cents 
were  paid  for  1  nutmeg;  and  what  should  be  charged  for  37 
nutmegs?  Ans.  $2.96. 

12.  Paid  2<£.  17s.  5d.  for  521bs.  of  sugar ;  what  cost  lib.  ? 
what  cost  761bs.  ?  Ans. 

13.  Paid  4JE.  3s.   lid.  for  76  pounds  of  sugar;  what  cost 
521bs.  ?  Ans. 

14.  If  521bs.  of  sugar  cost  2£.  17s.  5d.,  how  many  pounds 
can  be  purchased  for  4<£.  3s.  lid.  ?  Ans. 

15.  When  4<£.  3s.  lid.  are  paid  for  761bs.  of  sugar,  how 
many  pounds  can  be  obtained  for  2<£ .  17s.  5d.  ?     Ans. 


SECT,  xvi.]  VULGAR  FRACTIONS.  89 

16.  Bought  20  bushels  of  wheat  for  8£.  3s.  lid. ;  what  cost 
1  bushel  ?  what  cost  200  bushels  ?  Ans. 

17.  Paid  E.  Bradley  81£.  19s.  3d.  for  200  bushels  of  wheat ; 
what  cost  20  bushels  ?  Ans. 

18.  Mr.  Day  paid  3£.  4s.  2d.  for  10yds.  of  cloth  ;  what  should 
he  have  paid  for  97yds.  ?  Ans.  31£.  2s.  5d. 

19.  If  8  barrels  of  flour  cost  2£.  12s.,  what  cost  29  barrels  ? 

Ans.  9£.  8s.  6d. 

20.  If  17  bushels  of  wheat  cost  6£.  13s.  3d.,  what  cost  101 
bushels  ?  Ans.  39c£.  11s.  2d. 

21.  Gave  1Q£.  4s.  3d.  for  19  yards  of  cloth ;  what  cost  97 
yards  ?  Ans.  52 £.  2s.  9d. 


SECTION  XVI. 
VULGAR  FRACTIONS. 

FRACTIONS  are  parts  of  an  integer,  or  whole  number. 

An  integer  is  any  whole  number  or  quantity,  as  1,  7, 11,  &c., 
or  a  pound,  a  yard. 

VULGAR  FRACTIONS  are  expressed  by  two  numbers,  called 
the  Numerator  and  Denominator;  the  former  above,  and  the 
latter  below  a  line. 

rp,  C  Numerator     _7_ 

InUS,      £  Denominator  11 

The  Denominator  shows  into  how  many  parts  the  integer,  or 
whole  number,  is  divided. 

The  numerator  shows  how  many  of  these  parts  are  taken,  or 
expressed  by  the  fraction. 

1.  A  proper  fraction  is  one  whose  numerator  is  less  than  the 
denominator ;  as  y. 

2.  An  improper  fraction  is  one  whose  numerator  exceeds  or 
is  equal  to  the  denominator  ;  as  ||  or  f . 

3.  A  single  or  simple  fraction  consists  of  but  one  numerator 
and  one  denominator;  as  |-. 

4.  A  compound   fraction  is  a  fraction  of  a  fraction,  connected 
by  the  word  of;  as  %  of  £  of  f  of  f- . 

5.  A  mixed  number  is  an  integer  with  a  fraction  ;  as  7^T,  5f . 

6.  A  complex  fraction  is  a  fraction  having  a   fraction   or   a 
mixed  number  for  its  numerator  or  denominator,  or  both ;  as, 

7i     *    t    SA  1 

9?r'  y  77  11'  C      9J' 

8* 


90  VULGAR  FRACTIONS.  [SECT.  xvi. 

7.  The  terms  of  a  fraction  are  the  numerator  and  denominator ; 
the  numerator  being  the  upper  term,  and  the  denominator 
the  lower. 

8.  The  greatest  common  measure  of  two  or  more  numbers 
is  the  largest  number  that  will  divide  them  without  a  re- 
mainder. 

9.  The  least  common  multiple  of  two  or  more  numbers  is  the 
least  number  that  may  be  divided  by  them  without  a  re- 
mainder. 

10.  A  fraction  is  in  its  lowest  terms,  when  no  number  but  a 
unit  will  measure  both  its  terms. 

11.  A  prime  number  is  that  which  can  be  measured  only  by 
itself  or  a  unit;  as  7,  11,  and  19. 

12.  Numbers  are  said  to  be  prime  to  each  other,  when  only  a 
unit  measures  or  divides  them  both  without  a  remainder ; 
thus,  7  and  1 1  are  prime  to  each  other. 

13.  Prime  •  factors  of  numbers  are  those  factors  which  can  be 
divided  by  no  number  but  by  themselves  or  a  unit ;  thus  the 
prime  factors  of  21  are  7  and  3. 

14.  An  even  number  is  that  which  can  be  divided  into  two 
equal  whole  numbers. 

15.  An  odd  number  is  that  which  cannot  be  divided  into  two 
equal  whole  numbers. 

16.  A  square  number  is  the  product  of  a  number  multiplied  by 
itself. 

17.  A  cube  number  is  the  product  of  a  number  multiplied  by  its 
square. 

18.  A  composite  number  is  that  produced  by  multiplying  two 
or  more  numbers  together. 

19.  The  factors  of  a  number  are  those  whose  continued  prod- 
uct will  exactly  produce  the  number. 

20.  An  aliquot  part  is  that  which  is  contained  a  precise  num- 
ber of  times  in  another. 

21.  An  aliquant  part  is  such  a  number  as  is  contained  in  an- 
other a  certain  number  of  times  with  some  part  or  parts  over. 

22.  A  perfect  number  is  that  which  is  equal  to  the  sum  of  all 
its  aliquot  parts,  or  is  equal  to  the  sum  of  all  the  numbers 
that  will  divide  it  without  a  remainder ;  thus  6  is  a  perfect 
number,  because  it  can  be  divided  by  3,  2,  and  1 ;  and  the 
sum  of  these  numbers  is  6.     But  12  is  not  a  perfect  number, 
because  its  aliquot  parts  are  more  than  12 ;  thus  6  +  4  +  3  +  1 
—  14.     8  is  not  a  perfect  number,  because  its  aliquot  parts 
are  less  than  8 ;  thus  4  +  2+1  =  7.    But  28,  496,  and  8 128 


SECT,  xvi.]  VULGAR  FRACTIONS.  91 

are  perfect  numbers.     The  chief  use  of  a  knowledge  of  these 
numbers  is  in  the  higher  branches  of  mathematics. 

23.  A  fraction  is  equal  to  the  number  of  times  the  numerator 
will  contain  the  denominator. 

24.  The  value  of  a  fraction  depends  on  the  proportion  which 
the  numerator  bears  to  the  denominator. 

25.  Ratio  is  the  relation  which  two  numbers  or  quantities  of 
the  same  kind  bear  to  each  other,  and  may  be  found  by  divid- 
ing one  number  by  the  other.     For  example,  the  ratio  of  12 
to  4  is  3,  because  12  -i-  4  =  3  ;  and  the  ratio  of  4  to  8  is  £, 
because  4  -j-  by  8  =  £. 

CASE  I. 

To  find  the  greatest  common  measure  of  two  or  more  num- 
bers, or  to  find  the  greatest  number  that  will  divide  two  or  more 
numbers,  without  a  remainder. 

RULE.  —  Divide  the  greater  number  by  the  less,  and,  if  there  be  a  re- 
mainder, divide  the  last  divisor  by  it,  and  so  continue  dividing  the  last 
divisor  by  the  last  remainder  until  nothing  remains,  and  the  last  divisor 
is  the  greatest  common  measure. 

If  there  be  more  than  two  numbers,  find  the  greatest  common  measure 
of  two  of  them,  and  then  of  that  common  measure  and  the  other  numbers. 
If  it  should  happen  that  1  is  the  common  measure,  the  numbers  are  prime 
to  each  other,  and  are  incommeasurable. 

The  above  rule  may  be  illustrated  and  demonstrated  by  the 
following  example. 

Let  it  be  required  to  find  the  greatest  common  measure  or 
divisor  of  24  and  88. 

According  to  the  rule,  we  first  divide  88,  the  greater  num- 
ber, by  24,  the  less ;  for  it  is  evident  that  no  number  greater 
than  the  less  of  two  numbers  can  measure  or  divide  those 
numbers.  As  therefore  24  will  exactly  measure  or  divide  it- 
self, if  it  will  also  divide  88,  it  will  be  the  greatest  common  di- 
visor sought. 

Now  we  find  that  24  will  not  exactly  meas- 

OPERATION.  ,.    .,       00     ,      ^      -  .  J    .     j 

,  ure   or  divide  88,  but  there  is  a  remainder, 

70  16.     24,  therefore,  is  not  the  common  divisor 

_  of  the  two  numbers.     Now  as  72,  the  number 

16)24(1  which  we  subtracted  from  88,    is  an  exact 

16 multiple  of  24,  we  know  that  any  number 

8)16(2  which  will  exactly  measure  or  divide  24  will 

16  also  divide  72  ;  and  as  16,  the  remainder  of 

the  division  of  88  by  24,  is  that  part  of  88 


92  VULGAR  FRACTIONS.  [SECT.  xvi. 

which  24  will  not  measure  or  divide,  it  is  a  number  which 
must  be  divided  by  the  common  divisor  of  24  and  88.  Now, 
since  no  number  can  divide  16  greater  than  16  itself,  and 
since,  if  it  will  divide  24,  we  know  that  it  will  also  divide  88, 
because  88  is  a  multiple  of  24,  -{-  16,  we  proceed  according  to 
the  rule  to  try  whether  16  will  measure  or  divide  24,  and  there- 
fore place  24,  the  last  divisor,  at  the  right  of  16,  the  last  re- 
mainder. We  know,  also,  that  if  16  will  divide  24  it  is  the 
greatest  common  divisor  of  24  and  88  ;  because  we  have  before 
shown  that  any  number  which  will  divide  24  and  88  must  also 
divide  16. 

On  dividing  24  by  16  we  again  find  a  remainder,  8.  Now 
8  being  the  remainder  after  the  division  of  24  by  16,  we  know, 
according  to  the  reasoning  before  adopted,  that  no  number 
greater  than  8  can  measure  or  divide  16  and  24,  and  that  if  it 
will  measure  16,  it  will  also  measure  24,  because  24  is  a  mul- 
tiple of  16,  +  8,  and  that  for  the  same  reason  it  will  divide  88, 
for  88  is  a  multiple  of  24, -}- 16-  Making  8,  therefore,  the 
divisor,  and  16  the  dividend,  according  to  the  rule,  we  find  that 
8  will  exactly  divide  16,  and  hence  know  that  8  is  the  greatest 
common  divisor  of  24  and  88.  Q.  E.  D. 

2.  What  is  the  greatest  common  measure  of  56  and  168  ? 

Ans.  56. 

3.  What  is  the  greatest  common  measure  of  96  and  128  ? 

Ans.  32. 

4.  WThat  is  the  greatest  common  measure  of  57  and  285  ? 

Ans.  57. 

5.  What  is  the  greatest  common  measure  of  169  and  175  ? 

Ans.  1. 

6.  What  is  the  greatest  common  measure  of  175  and  455  ? 

Ans.  35. 

7.  What  is  the  greatest  common  measure  of  169  and  866  ? 

Ans.  1. 

8.  What  is  the  greatest  common  measure  of  47  and  478  ? 

Ans.  1. 

9.  What  is  the  greatest  common  measure  of  84  and  1068  ? 

Ans.  12. 

10.  What  is  the  greatest  common  measure  of  75  and  165  ? 

Ans.  15. 

11.  What  is  the  greatest  common  measure  of  78,  234,  and 
468  ?  Ans.  78. 


SECT,  xvi.]  VULGAR  FRACTIONS.  93 

12.  What  is  the  greatest  common  measure  of  144,  485,  and 
25  ?  Ans.  1. 

13.  What  is  the  greatest  common  measure  of  671,  2013,  and 
4026?  Ans.  671. 

14.  What  is  the  greatest  common  measure  of  16,  20,  and 
24  ?  Ans.  4. 

15.  What  is  the  greatest  common  measure  of  21,  27,  and  81  ? 

Ans.  3. 

CASE  II. 

To  reduce  fractions  to  their  lowest  terms. 
1.  Reduce  £f  to  its  lowest  terms. 

OPERATION. 


NOTE.  —  That  J  is  equal  to  ||  may  be  demonstrated  as  follows  :  —  16  is 
the  same  multiple  of  1,  that  48  is  of  3,  therefore  16  has  the  same  ratio  to 
48,  that  1  has  to  3  ;  and  as  the  value  of  a  fraction  depends  on  the  ratio 
which  the  numerator  has  to  the  denominator,  it  is  evident  when  their 
ratios  are  the  same  that  their  values  are  equal  ;  therefore,  J  is  equal  to  i| 

Q.  E.  D. 

RULE.  —  Divide  the  numerator  and  denominator  by  any  number,  that 
will  divide  them  both  ivithout  a  remainder;  and  so  continue,  until  no 
number  ivill  divide  them  but  a  unit.  Or  divide  the  numerator  and  de- 
nominator by  their  greatest  common  measure. 

2.  Reduce  ^f  to  its  lowest  terms.  Ans.  /T. 

3.  Reduce  §|  to  its  lowest  terms.  Ans.  f  . 

4.  Reduce  f  £  to  its  lowest  terms.  Ans.  f  . 

5.  Reduce  •££%  to  its  lowest  terms.  Ans.  -^¥. 

6.  Reduce  -j^  to  its  lowest  terms.  Ans.  -£$. 

7.  Reduce  £f  f  to  its  lowest  terms.  Ans.  -f^f 

8.  Reduce  ^VrV  to  i*s  lowest  terms.  Ans.  ^. 

9.  Reduce  ||f  to  its  lowest  terms.  Ans.  f|f  . 

10.  Reduce  -fflt  to  l^  lowest  terms.  Ans.  T 

11.  Reduce  f-j-f  to  its  lowest  terms.  Ans. 

12.  Reduce  TVT%  to  its  lowest  terms.  Ans. 

13.  Reduce  £f  f  to  its  lowest  terms.  Ans. 

CASE  III. 

To  reduce  mixed  numbers  to  improper  fractions. 
1.  How  many  fifths  of  a  gallon  in  17f  gallons  ? 


94  VULGAR  FRACTIONS.  [SECT.  xvi. 

OPERATION.  We  analyze  this  question  by  saying,  that,  as 

I<j3_  there  are  5  fifths  in   1   gallon,   there  will  be  5 

5&  times  as  many  fifths  as  gallons.     Therefore  in 

aa  _     A  17  gallons  and  3  fifths  of  a  gallon  there  will  be 

~  Ans'  88  fifths,  which  should  be  expressed  thus,  \*. 

And  this  fraction  by  definition  2d,  page  89,  is  an  improper 

fraction. 

RULE.  —  Multiply  the  whole  number  by  tlie  denominator  of  the  frac- 
tion, and  to  the  product  add  the  numerator,  and  place  their  sum  over  the 
denominator  of  t  tie  fraction. 

2.  Reduce  16T3T  to  an  improper  fraction.  Ans. 

3.  Reduce  14f  to  an  improper  fraction.  Ans. 

4.  Reduce  126£f  to  an  improper  fraction.  Ans. 

5.  Reduce  149|£f  to  an  improper  fraction.  Ans. 

6.  Reduce  16  1-^  to  an  improper  fraction.  Ans. 

7.  Reduce  171f£  to  an  improper  fraction.  Ans.  - 

8.  Reduce  98|§  to  an  improper  fraction.  Ans. 

9.  Reduce  1  1  6£|  to  an  improper  fraction.  Ans. 

10.  Reduce  718ff  to  an  improper  fraction.      Ans. 

11.  Reduce  100|£g  to  an  improper  fraction.  Ans.  * 

12.  Reduce  478y^F  to  an  improper  fraction.    Ans.  ^ 

13.  Reduce  87  1T|F  to  an  improper  fraction.  Ans. 

14.  Reduce  167^4r  to  an  improper  fraction.    Ans. 

15.  Reduce  613  j^  to  an  improper  fraction.  Ans. 

16.  Reduce  159£g£  to  an  improper  fraction.    Ans. 

17.  Reduce  999T9^9?  to  an  improper  fraction.  Ans. 

18.  Reduce  7  to  an  improper  fraction.  Ans. 

19.  Change  11  to  an  improper  fraction.  Ans. 

20.  Change  1  to  an  improper  fraction.  Ans. 

21.  Change  100  to  an  improper  fraction.  Ans. 


NOTE.  —  To  reduce  a  whole  number  to  an  equivalent  fraction  having  a 
given  denominator,  we  multiply  the  whole  number  by  the  given  denom- 
inator, and  we  then  place  the  product  over  the  given  denominator. 

22.  Reduce  11  to  a  fraction  whose  denominator  shall  be  7. 

OPERATION. 

11  X  7  =  77;  V  Ans. 

23.  Reduce  5  to  a  fraction  whose  denominator  shall  be  17. 

Ans.  ff 

24.  .Reduce  19  to  a  fraction  whose  denominator  shall  be  13. 

Ans. 


SECT,  xvi.]  VULGAR  FRACTIONS.  95 

CASE  IV. 

To  reduce  improper  fractions  to  integers  or  mixed  numbers. 

1.  How  many  yards  in  yff7  of  a  yard  ? 

OPERATION.  This  question  may  be  analyzed  by  say- 

19)117(6-3   Ans      *n£'  as  ^  nineteenths  make  one  yard,  there 

114     T  w^  ke  as  many  yards  as  117  contains  19, 

—  which  is  6  times  and  3  nineteenths  times, 

which  is  written   thus,  6T3g- ;  and  this   ex- 
pression, by  definition  5th,  page  89,  is  a  mixed  number. 

RULE.  — Divide  the  numerator  by  the  denominator,  and  if  there  be  a 
remainder,  place  it  over  the  denominator  at  the  right  hand  of  the  integer. 

2.  Reduce  *fg  to  a  mixed  number.  Ans. 

3.  Reduce  ^f^1  to  a  mixed  number.  Ans. 

4.  Reduce  ^3Tl  to  a  mixed  number.  Ans.  7-ff . 

5.  Reduce  f  £]-  to  a  mixed  number.  Ans.  3[§f. 

6.  Change  ^-^-^  to  a  mixed  number.  Ans.   111^. 

7.  Change  ^f-  to  a  mixed  number.  Ans.  9 Iff. 

8.  Change  J-f-5-  to  a  whole  number.  Ans.   125. 

9.  Change  ff  to  a  whole  number.  Ans.  1. 

CASE  V. 

To  reduce  complex  fractions  to  simple  ones. 

2 

1.  Reduce  -  to  a  simple  fraction. 

OPERATION  ^n  Perf°rming  tnis  question,  we  divide 

2  the  numerator  by  the  denominator ;  be- 

|  =  f  X  f  =  if  Ans.  cause  all  fractions  are  equal  to  the  num- 
ber of  times  the  numerator  contains  the 
denominator. 

RULE.  — If  the  numerator  or  denominator  be  a  whole  or  a  mixed  num- 
ber, let  it  be  reduced  to  an  improper  fraction.  Then  multiply  the  denom- 
inator oftfie  lower  fraction  into  the  numerator  of  the  upper  fraction  for  a 
new  numerator,  and  the  denominator  of  the  upper  fraction  into  the  numer- 
ator of  the  lower  fraction  for  a  new  denominator  ;  or,  we  may  invert  the 
denominator  of  the  complex  fraction,  when  reduced,  and  place  it  in  a  line 
with  the  numerator,  then  multiply  the  two  numerators  together  for  a  new 
numerator,  and  the  denominators  together  for  a  new  denominator. 

NOTE.  —  Every  fraction  denotes  a  division  of  the  numerator  by  the 
denominator,  and  its  value  is  equal  to  the  quotient  obtained  by  such  divis- 
ion. Hence  the  necessity  of  inverting  the  denominator  of  the  complex 
fraction. 


96  VULGAR  FRACTIONS.  [SECT.  xvi. 

2.  Reduce  -  to  a  simple  fraction. 

15 

OPERATION. 

£  =  \  x  J  =  V  =  10£  Ans. 

3 
3. 

3.  Reduce  £  to  a  simple  fraction. 

o 

OPERATION. 


43 

4.  Reduce  -^  to  a  simple  fraction. 


OPERATION. 


3 

5.  Reduce  ^  to  a  simple  fraction. 


OPERATION. 


7 
6.  Reduce  —  to  a  simple  fraction. 

OPERATION. 


7  * 
7.  Reduce  -^  to  a  simple  fraction. 

o 

OPERATION. 


8.  Reduce  r    to  a  simple  fraction. 
of 

OPERATION. 

?f  —   ^  =   5  6    vx      3     —    168 
go  -   OB  -     »  26  234 


SECT.  xvi.J  VULGAR  FRACTIONS.  97 

2 

9.  Reduce  |  to  a  simple  fraction.  Ans.  J£. 

Q 

10.  Reduce  -  to  a  whole  number.  Ans.  24. 

IT 

11.  Reduce  |  to  a  simple  fraction.  Ans.  f. 

12.  Reduce  ~  to  a  mixed  number.  Ans.  12|. 

13.  Reduce  ~  to  a  simple  fraction.  Ans.  •£$. 

o 

14.  Reduce  —  to  a  mixed  number.  Ans.  1£. 

31. 

15.  Reduce  -^  to  a  simple  fraction.  Ans.  4f  . 

y 

ll£ 

16.  Reduce  j^|  to  a  simple  fraction.  Ans.  f  f  . 

17.  If  7  were  to  be  the  denominator  to  the  following  quantity, 

7£ 
•—  j^,  what  would  be  its  value  ?  Ans.  -^y. 

CASE  VI. 

To  reduce  compound  fractions  to  simple  fractions. 
1.  What  is  £  of  £? 

OPERATION.  This  question  may  be  analyzed  by  saying, 

s  x  Z.—  :  24.  Ans.  ^i  °f  an  orange  be  divided  into  4  equal 
parts,  one  of  those  parts  is  -^  of  the  orange  ; 
and,  if  £  of  £  be  ^-,  it  is  evident  that  £  of  £  will  be  seven  times 
as  much.  And  7  times  ^  ig  ife-  If»  therefore,  ^-  of  |-  be  g^-,  J 
of  |-  will  be  3  times  as  much  ;  and  3  times  -^  is  f  £  . 

RULE.  —  Change  mixed  numbers  and  whole  numbers,  if  there  be  any, 
to  improper  fractions  ;  then  multiply  all  the  numerators  together  for  a 
new  numerator,  and  all  the  denominators  together  for  a  new  denominator  ; 
the  fraction  should  then  be  reduced  to  its  lowest  terms.  If  there  be  num- 
bers in  the  numerator  similar  to  those  in  the  denominator,  they  may  be 
cancelled  in  the  operation. 


2.  What  is  §  of  £  of  £  of  •&  ?  Ans.  ^T  =  f  II- 

3.  What  is  f  of  -B  of  f  of  TV  ?  Ans. 


98  VULGAR  FRACTIONS.  [SECT.  xvi. 


4.  What  is  f  of  |  of  J  of  {£  ?  Ans.  ^jft  = 

5.  What  is  T6T  of  f  of  £  of  21  ?  Ans.  £$£  =  2T5T- 

6.  What  is  tV  of  15£  of  5^  of  100  ?  Ans.  *af  J§att  =  5T58H, 

7.  What  is  }  off  of  T7T  o 


STATEMENT.  CANCELLED. 

3  4        7       11       3       *„   1  „**        3        !  . 

4  X  7  X  UX  24  =  i  X  *  X  H  X  24=24  =  8  Ans' 

8.  What  is  f  of  TV  of  Jf  of  ff  ? 

STATEMENT.  CANCELLED. 

3vI_v^v2?-3v^y£^x^=-A--l  An 
7X15     29  X33"~^fA^     &$     33       33       11  ^ 

9.  What  is  .ft  of  i|  of  J  J  of  f  |  ? 

STATEMENT.  CANCELLED. 

7  v11v:LIv2?-  Zv£fv£?v^—  1   -!A 
11X17     23X28~"i:a:     ^X^     28"~28""4A 

10.  What  is  £  of  §  of  f  of  £  of  tV  of  24  ? 


11.  What  is  the  value  of  T7T  of  £|  of  f  f  of  $  7}  ? 
7      ^^^7 


12.  What  is  |  of  T9T  of  T  J  of  3f  gallons  ? 


RULE  2.  —  When  tJiere  are  any  two  numbers,  one  in  the  numerators, 
and  the  other  in  the  denominators,  which  may  be  divided  by  a  number 
without  a  remainder,  the  quotients  arising  from  such  division  may  be 
used  in  the  operation  of  the  question,  instead  of  the  original  numbers. 
The  quotients  also  may  be  cancelled,  as  other  numbers. 


1.  Reduce  f  of  ^-f  of  %\  of  T5T  to  its  lowest  terms. 

OPERATION.  In  performing  this  question, 

271  we  find  that  14  among  the  nu- 

_  56  merators,  and  7  among  the  de- 

~495  AnS*   nominators,  may  be  divided  by 
7,  and  that  their  quotients  will 


SECT,  xvi.]  VULGAR  FRACTIONS.  99 

be  2  and  1.  We  write  the  2  above  the  14,  and  1  below  the  7. 
We  also  find  a  21  among  the  numerators,  and  a  27  among  the 
denominators,  which  may  be  divided  by  3,  and  that  their  quo- 
tients will  be  7  and  9.  We  write  the  7  above  the  21,  and  9  be- 
low the  27.  We  again  find  a  5  among  the  numerators,  and  a 
25  among  the  denominators,  which  may  be  divided  by  5,  and 
that  their  quotients  will  be  1  and  5.  We  write  the  1  over  the 
5,  and  the  5  below  the  25.  We  then  multiply  the  4,  2,  7,  and 
1  together  for  a  numerator  =  56,  and  the  1,  9,  5,  and  11  for  a 
denominator  =  495.  The  answer  will  therefore  be 


2.  Reduce  if  of  £f  of  J-f  of  /T  to  a  simple  fraction. 
2621 

24 

-~         Ans' 


55  # 

3.  What  is  the  value  of  f  of  &  of  -ff  of  |f  of  $  34  ? 
1  3      g      g 


NOTE.  —  The  above  rule  will  apply,  when  the  product  of  several  num- 
bers is  to  be  divided  by  the  product  of  other  numbers. 

4.  What  is  the  continued  product  of  8,  4,  9,  2,  12,  16,  and  5, 
divided  by  the  continued  product  of  40,  6,  6,  3,  8,  4,  and  20  ? 

1 

i 

="  5  AllS' 


5 

The  product  of  4  and  9  in  the  upper  line  is  equal  to  the 
product  of  6  and  6  in  the  lower,  therefore  they  are  cancelled  ; 
and  the  product  of  2  and  12  in  the  upper  line  is  equal  to  the 
product  of  3  and  8  in  the  lower  line  ;  also  the  product  of  16  and 
5  in  the  upper  line  is  equal  to  the  product  of  4  and  20  in  the 
lower  line  ;  these  are  all  cancelled.  We  also  find,  that  the  8  in 
the  upper  line  and  the  40  in  the  lower  line  may  be  divided  by 
8,  and  their  quotients  will  be  1  and  5.  We  write  the  1  above 
the  8,  and  the  5  below  the  40.  By  the  usual  process,  we  now 
find  our  answer  is  •. 


100  VULGAR  FRACTIONS.  [SECT.  ivi. 

5.  What  is  the  continued  product  of  12,  13,  14,  15,  16,  18, 
20,  21,  and  24,  divided  by  the  continued  product  of  2,  3,  4,  5, 
6,7,8,9,  10,  and  11? 

3  2322272 


11111111    i 

CASE  VII. 

To  find  the  least  common  multiple  of  two  or  more  numbers. 

1.  What  is  the  least  common  multiple  of  4,  6,  8,  16,  and  20  ? 

OPERATION.  In   tms   operation,   we    have 

2H     6     8     16     20  divided  the   given  numbers  by 

i  ---  that  number  which  would  divide 

2)2     3     4       8     10  most  Of  tne  given  numbers  with- 

2)1     3     2       4       5  out  a  remainder,  that  is,  by  one 

"l     3     i       2       5  of  the  prime  factors,  and  have 

—  24ft  An*   continued  the  division  until  no 
0  Ans. 


them.     We  have  then  multiplied  all  the  divisors  and  the 
quotients  together  and  found  their  product  to  be  240,  which  is  a 
common  multiple  of  the  given  numbers,  4,  6,  8,  16,  and  20. 

It  is  not  only  a  common  multiple,  but  it  is  the  least  common 
multiple. 

To  prove  this,  we  assume  the  following  admitted  propositions, 
which  are  self-evident  :  — 

1.  Every  number  not  prime  itself  is  the  product  of  two  or 
more  primes  or  factors,  and  is  resolvable  into  its  original  primes 
by  division. 

2.  The  least  number  which  contains  all  the  prime  factors  of 
two  or  more  numbers  is  the  least  common  multiple  of  those 
quantities. 

The  factors  of  each  number  in  the  question  are  as  follows  :  — 
Factors  of  4    are          2  and  2  =  2  X  2  =  4 
6     "  2   and3:=2'X3  =  6 

8     "      2,  2,  and  2  =  2  X  2  X  2  =  8 
16     "2,2,2,  and  2  =  2x2x2  X2=16 
20     "      2,  2,  and  5  =  2  x  2  x  5  =  20 

Now  240  is  the  least  number  which  contains  all  the  factors 
common  to  each  of  these  numbers,  2x2x2x2x3x5  = 


SECT,  xvi.]  VULGAR  FRACTIONS.  101 

240  ;  therefore  it  is  the  least  common  multiple  of  all  these 
numbers. 

Or  the  above  principle  may  be  illustrated  as  in  the  following 
question. 

2.  What  is  the  least  common  multiple  of  4, 5,  6, 7,  8,  and  9  ? 
2)4    5     6    7    8    9 


2)2     5     3     7     4     9 


In  this  question,  we  divide   by   the 


2)1     5  729  prime  factors,  2,  3,  5,  7,  &c.,  succes- 

3)153719  sively,  until  the  last  quotients  terminate 

3  jl 5     I     7     T~1J  m  units  5  and    tne   product   of  all   the 

: — ~ — _ —   •  prime  factors  is  the  least  common  mul- 

5)*     5     *     7     *     *  tiple. 
7)1     1     1     7     1     1 

111111 
Thus2x2x2x3x3  X  5x7  =  2520  Ans. 

That  2520  is  the  least  common  multiple  is  evident  from  the 
fact,  that  the  divisors  whose  product  produces  this  number  are 
the  prime  factors,  and  the  only  prime  factors  of  the  given  num- 
bers, as  may  be  seen  below. 

Factors  of  4  are          2  and  2  =  2  x  2  =  4 

5  "  5  and  1  =  5  x  1  =  5 

6  "  2  and  3  ==  2  x  3  =  6 

7  "  7  and  1  =  7  x  1  =  7 

8  "  2,  2,  and  2  =  2  x  2  x  2  =  8 

9  "  3  and  3  =  3  X  3  =  9 

RULE.  —  Divide  by  such  a  number  as  will  divide  most  of  the  given 
numbers  without  a  remainder,  and  set  the  several  quotients ,  with  the 
several  undivided  numbers,  in  a  line  beneath,  and  so  continue  to  divide, 
until  no  number  greater  than  unity  will  divide  two  or  more  of  them. 
Then  multiply  all  the  divisors,  the  last  quotients,  and  undivided  num- 
bers together,  and  the  product  is  the  least  common  multiple. 

Or,  divide  the  given  numbers  successively  by  their  prime  factors,  until 
the  last  quotients  terminate  in  units.  The  continued  product  of  all  the 
divisors  will  be  the  least  common  multiple. 

3.  What  is  the  least  common  multiple  of  6,  8,  10,  18,  20, 
and  24  ?  Ans.  360. 

4.  What  is  the  least  common  multiple  of  14,  19,  38,  and  57  ? 

Ans.  798. 

5.  What  is  the  least  common  multiple  of  20,  36,  48,  and 
50  ?  Ans.  3600. 

9* 


102  VULGAR  FRACTIONS.  [SECT.  xvi. 

6.  What  is  the  least  common  multiple  of  15,  25,  35,  45,  and 
100  ?  Ans.  6300. 

7.  What  is  the  least  common  multiple  of  100,  200,  300,  400, 
and  575  ?  Ans.  27600. 

8.  I  have  four  different  measures  ;  the  first  contains  4  quarts, 
the  second  6  quarts,  the  third  10  quarts,  and  the  fourth   12 
quarts.     How  large  is  a  vessel,  that  may  be  filled  by  each  one 
of  these,  taken  any  number  of  times  full  ?       Ans.  60  quarts. 

NOTE.  —  In  finding  the  common  multiple  of  two  or  more  numbers,  any 
one  number  that  can  measure  another  may  be  cancelled. 

9.  What  is  the  least  common  multiple  of  4,  6,  8,  12,  16,  10, 
and  20  ? 

2°     4X3X4X5  =  240  Ans. 


34 

By  examining  this  question,  we  find  that  8  may  be  divided 
by  4,  12  by  6,  16  by  8,  and  20  by  10  ;  therefore  we  cancel  4, 
6,  8,  and  10. 

10.  What  is  the  least  common  multiple  of  5,  15,  30,  7,  14, 

and  28  ? 

tf  30  y  U  28    2xl5xl4=420An, 


15  14 

In  this  question,  we  find  that  15  may  be  measured  by  5,  30 
by  15,  14  by  7,  and  28  by  14  ;  we  therefore  cancel  5,  15,  7, 
and  14. 

11.  What  is  the  least  common  multiple  of  1,  2,  3,  4,  5,  6,  7, 
8,  and  9  ? 

2)*  *  JM  56789    2X3X5X7X4X3  =  2520 
~3)5  3749  tAns' 

51743 

12.  What  is  the  least  common  multiple  of  9,  8,  12,  18,  24, 
36,  and  72  ? 

0  0  *«  *0  U  30  72        72  Ans. 

13.  What  is  the  least  number  that  18,  24,  36,  12,  6,20,  and 

48  will  measure  ? 


SECT,  xvi.]  VULGAR  FRACTIONS.  103 

4)  t$  #4  36  a#  0  20  48     4X3X3X5X4  =  720  Ans. 
3)9  5  12 

IT"  5     4 

,  CASE  VIII. 

To  reduce  fractions  to  a  common  denominator,  that  is,  to 
change  fractions  to  other  fractions,  all  having  their  denomina- 
tors alike,  yet  retaining  the  same  value. 

1.  Reduce  £,  T52-,  and  ±fa  to  other  fractions  of  equal  value, 
having  the  same,  or  a  common,  denominator. 

First  Method. 

OPERATION. 

4)8  12  16    4X2X3X2=48  common  denominator. 


2)2     34  8 


132 


6  X  7=42  numerator  for  £•  =f  f . 
4  X  5=20  numerator  for  •&•=§$• 
3  X 1 1=33  numerator  for  H=$ f . 


Having  first  obtained  the  least  common  multiple  of  all  the 
denominators  of  the  given  fractions  by  the  last  rule,  we  assume 
this  as  the  common  denominator  required.  This  number  (48) 
we  divide  by  the  denominators  of  the  given  fractions,  8,  12, 
and  16,  and  find  their  quotients  to  be  6,  4,  and  3,  which  we 
place  under  the  48  ;  these  numbers  we  then  multiply  by  the 
numerators  7,  5,  and  11,  and  find  their  products  to  be  42,  20, 
and  33,  and  these  numbers  are  the  numerators  of  the  fractions 
required. 

Second  Method. 

OPERATION. 

7  X  12  X  16  =  1344  numerator  for    J-  =  |f  £f. 
5  X    8  X  16  =    640  numerator  for  T^  = 
11  X    8  X  12  =  1056  numerator  for  J£  = 


The  numerators  are  produced  by  multiplying  the  numer- 
ators of  the  given  fraction  by  each  of  the  other  denomina- 
tors, and  the  common  denominator  is  obtained  by  multiplying 
all  the  denominators.  By  this  process,  we  obtain  the  follow- 
ing  ;  Ans.  ff^,  T*ftfr,  iMf  • 

The  pupil  will  perceive,  that  this  method  does  not  express 
the  fractions  in  so  low  terms  as  the  other  ;  although  they  both 
have  the  same  value. 


104  VULGAR  FRACTIONS.  [SECT.  xvi. 

RULE.  —  Find  the  least  common  multiple  of  all  the  denominators  by 
Case  VII.  ,  and  it  will  be  the  denominator  required.  Divide  the  com- 
mon multiple  by  each  of  the  denominators,  and  multiply  the  quotients  by 
the  respective  numerators  of  the  fractions  and  their  products  will  be  the 
numerators  required. 

Or,  multiply  each  numerator  into  all  the  denominators  except  its  own 
for  a  new  numerator  ;  and  all  the  denominators  into  each  other  for  a 
common  denominator. 

Questions  to  be  performed  by  the  first  method. 

2.  Reduce  f  ,  f  ,  &,  and  A-  Ans.  f  f  ,  f  °,  f  },  ££. 

3.  Reduce  A.  A,«,and  A-  Ans.  # 

4.  Reduce  |,  A,  A.  and  ^  Ans.  ££,  |f  ,  ff  , 

5.  Reduce  £f  ,  *,  it.  and  £.  Ans.  f  J,  f  £,  f  f  ,  f  f  . 

6.  Reduce  J£,  T3T,  f  ,  and  f  .  Ans.  §£§,  „«&,  £$|,  §§£. 

7.  Reduce  f  ,  |,  f  ,  and  f  Ans.  |ft,  Jfc  |£,  f  J. 

8.  Reduce  f,  f,  A,  and  A-  Ans.  f  f  J,  «§,  |^,  J|J. 

9.  Reduce  f  ,  J,  i,  and  3|.  Ans.  ^,  |J^,  ^,  *-£§. 

10.  Reduce  ^,  f  ,  *,  and  4J.       Ans.  TV¥,  T^F,  ^F,  1§|. 

11.  Reduce  f,  T5¥,  ^|,  and  f  Ans.  ||,  ^,  f  j,  ff 

12.  Reduce  f  ,  -&,  |J,  and  ?\.     Ans.  f  ff,  ^^J,  f|f  ,  ^fc. 

13.  Reduce  J§,  ^,  i^,  and  ^.  Ans.  |Jfc,  f^,  ^§^,  FVV 

14.  Reduce  J,  ^,  T\,  and  ^.    Ans.  f^g, 

15.  Reduce  f,  7,  8,  and  5f  .          Ans.  if, 

Questions  to  be  performed  by  the  second  method. 

16.  Reduce  f  ,  f  ,  and  £  to  fractions  having  a  common  de- 
nominator. Ans. 

17.  Reduce  f  ,  f  ,  and  ft.  Ans. 

18.  Reduce  T6T,  f  ,  and  TV  Ans. 

19.  Reduce  ^,  ft,  and  7|.  Ans. 

20.  Reduce  j|,  |,  and  ^.  Ans.  ££  |f  ,  Ji 

21.  Reduce  f,  T4T,  and  11T\.        Ans. 

22.  Reduce  \,  f,  f  ,  and  8.  Ans.  f  4-,  ff  ,  ff, 

23.  Reduce  f,  ^r.  and  f  of  7|.     Ans.  ^Wi  A 

24.  Reduce  J,  f  ,  £,  and  17.  Ans.  f  |,  f  f  , 

25.  Reduce  |£,  A  Of  6,  and  21£.       Ans.  |^,  f 

26.  Reduce  f  ,  T4T,  T57,  f  ,  and  f 

Ans.  if  Slf 

27.  Reduce  T%,  £ft,  and 

Ans.  ^WtfftHSr.  AWWftSr. 


NOTE.  1.  —  If  there  be  complex  fractions,  they  must  first  be  changed 
to  simple  fractions,  and  then  they  may  be  reduced  by  the  foregoing 
Rule. 


SECT,  xvi.]  VULGAR  FRACTIONS.  105 

4    74-  3f 

28.  Reduce  —  ,  —  ,  and  —~  to  a  common  denominator. 


29.  Reduce  —  —  ,  —  ^,  and  —  J  to  a  common  denominator. 

11^     ol^  OOf- 

Ans.  A,  it,  M- 

g4 

30.  Reduce  16,  f  o^-jyV'  and  t  of  8t  to  a  common  de- 
nominator. Ans. 


NOTE  2.  —  To  reduce  a  whole  number  to  an  equivalent  fraction  hav- 
ing a  given  denominator,  we  multiply  the  whole  number  by  the  given 
denominator,  and  place  the  product  over  it. 

31.  Reduce  7  to  a  fraction  whose  denominator  shall  be  11. 

OPERATION. 

7X  11  =  77.     ft  Ans. 

32.  Reduce  18  to  a  fraction  whose  denominator  shall  be 
14.  Ans.  2f£-. 

33.  Reduce  100  to  a  fraction  whose  denominator  shall  be 
19.  Ans.  J-f  {p. 

34.  Reduce  53  to  a  fraction  whose  denominator  shall  be  9. 

Ans.  ±  j*. 
CASE  IX. 

To  reduce  fractions  of  a  lower  denomination  to  a  higher. 
1.  What  part  of  a  pound  is  f  of  a  penny  ? 

OPERATION.  This  question  may  be  thus  ana- 

^2  x    3.d.  =  ^    —  ^St  lyzed.     Since  12  pence  make  a  shil- 

i    v    u«  i    f  ^ng>  there  will  be  TV  as  many  shil- 

lings as  pence  ;  therefore  TV  of  f  of 

a  penny  is  ¥3¥  =  -fa  of  a  shilling.    Again,  as  20  shillings  make 
a  pound,  there  will  be  ^V  as  many  pounds  as  shillings  ;  there- 
fore ^  of  <£%  of  a  shilling  is  ^^  of  a  pound.     Q.  E.  D. 
This  question  may  be  abridged. 

Thus,  f  X  A  X  A  =  Tinnr  =  *iv  Ans- 


RULE.  —  Multiply  the  denominator  of  the  given  fraction  by  all  tJte 
denominations  between  it  and  tlie  one  to  which  it  is  to  be  reduced,  and 
over  the  product  write  the  given  numerator. 

2.  Reduce  f  of  a  farthing  to  the  fraction  of  a  pound. 

3.  Reduce  f  of  a  grain  troy  to  the  fraction  of  a  pound. 


106  VULGAR  FRACTIONS.  [SECT.  XTI. 

4.  Reduce  £  of  a  scruple  to  the  fraction  of  a  pound. 

5.  Reduce  T6T  of  an  ounce  to  the  fraction  of  a  hundred 
weight. 

6.  Reduce  £  of  a  pound  to  the  fraction  of  a  ton. 

7.  Reduce  £  of  an  inch  to  the  fraction  of  an  ell  English. 

8.  Reduce  f  of  an  inch  to  the  fraction  of  a  mile. 

9.  Reduce  £  of  a  barleycorn  to  the  fraction  of  a  league. 

10.  Reduce  £  of  an  inch  to  the  fraction  of  an  acre. 

11.  Reduce  £  of  a  quart  to  the  fraction  of  a  tun,  wine  meas- 
ure. 

12.  Reduce  f  of  a  pint  to  the  fraction  of  a  bushel. 

13.  Reduce  £  of  a  minute  to  the  fraction  of  a  year  (365J 
days). 

CASE  X. 

To  reduce  fractions  of  a  higher  denomination  to  a  lower. 
1.  What  part  of  a  penny  is  •$%-$  of  a  pound  ? 

OPERATION.  We  explain  this  question  in 

c£n  X  ¥  =  &°TF  =  i&s.  the    following    manner.      As 

i     v  j_2  -     J.2   -      3r\    A  no     shillings  are  twentieths  of  a 
w  X  ¥  -  |d.  Ans.  *  . 


as  many  parts  of  a  shilling  in  ^^  of  a  pound,  as  there  are 
parts  of  a  pound  ;  therefore,  ^^  of  a  pound  is  equal  to  ^^  of 
•^  =  -^j  =  1^7  of  a  shilling.  And  as  pence  are  twelfths  of 
a  shilling,  there  will  be  twelve  times  as  many  parts  of  a  penny 
in  -£2  of  a  shilling,  as  there  are  parts  of  a  shilling  ;  therefore, 
^2-  of  a  shilling  is  equal  to  -£$  of  -^  =  £f  =  £  of  a  penny, 
Ans. 

The  operation  of  this  question  may  be  facilitated  in  the  fol- 
lowing manner. 

OPERATION. 
X  -2T°-  X  V  =  f  f  g  =  f  d.  Ans. 


RULE.  —  Let  the  given  numerator  be  multiplied  by  all'  the  denomi- 
nations between  it  and  tJie  one  to  which  it  is  to  be  reduced  ;  then  place 
the  product  over  the  given  denominator  and  reduce  the  fraction  to  its 
lowest  terms. 

2.  Reduce  12T00  of  a  pound  to  the  fraction  of  a  farthing. 

3.  Reduce  96*00  of  a  pound  troy  to  the  fraction  of  a  grain. 

4.  Reduce  7^7  of  a  pound,  apothecaries'  weight,  to  the 
fraction  of  a  scruple. 

5.  Reduce  Tv\v  of  a  cwt.  to  the  fraction  of  an  ounce. 

6.  Reduce  p^jpj  °f  a  ton  to  tne  fraction  of  a  pound. 


SECT,  xvi.]  VULGAR  FRACTIONS.  107 

7.  Reduce  3^-5-  of  an  ell  English  to  the  fraction  of  an  inch. 

8.  Reduce  Tr^w  of  a  mile  to  the  fraction  of  an  inch. 

9.  Reduce  TTT^FU  °f  a  league  to  the  fraction  of  a  barley- 
corn. 

10.  Reduce  srtr/irsvv  °f  an  acre  to  tne  fraction  of  an  inch. 

11.  Reduce  yrVs  °f  a  tun  °f  wine  measure  to  the  fraction 
of  a  quart. 

12.  Reduce  ^f  -$  of  a  bushel  to  the  fraction  of  a  pint. 

13.  Reduce  i^rfirsTr  °f  a  year  to  tne  fraction  of  a  minute. 

CASE  XI. 

To  find  the  value  of  a  fraction  in  the  known  parts  of  the 
integer. 

1.  What  is  the  value  of  T3T  of  a  <£.  ? 

By  Case  IX.  •&£.  =  ^.X^-=  f  ?s.  =  5T5Ts.  ;  and  T\s.  =. 
r5T  X  V2  =  «d.  =  5&d.  ;  and  T*Td.  -  T5T  X  *  -  ftqr.  = 


This  question  may  be  analyzed  thus:  —  If  l£.  is  20s.,  T3T 
of  a  £.  is  T3T  of  20s.  =  5T5Ts.  ;  and  if  Is.  is  12d.,  T5T  of  a  shil- 
ling is  T5T  of  12d.  =  5T5Td.  ;  and  if  Id.  is  4qr.,  T5T  of  a  penny  is 
T5T  of  4qr.  =  1-^j-qr.,  Answer,  as  before. 

Or,  it  may  be  performed  in  the  following  manner. 

OPERATION. 

In   this  operation,  it  will   be   per- 

ceived that  we  multiply  the  numer- 

11)60  (5s.  ator  °f  tne  fraction  by  the  successive 

55  lower  denominations,  beginning  with 

—g  the  highest,  and  divide  each  product 

j2  by  the  denominator. 

ll)"60(5d. 
j>5 
5 
_4 

ll)20(lT9Tqr.  From  these  illustrations,  we  derive 

the  following 
9 

RULE.  —  Multiply  the  numerator  by  the  next  lower  denomination  of 
the  integer,  and  divide  tJte  product  by  the  denominator  ;  if  any  thing 
remain,  multiply  it  by  the  next  less  denomination,  and  divide  as  before, 
and  so  continue  as  far  as  may  be  required;  and  the  several  quotients 
will  be  the  answer. 


108  VULGAR  FRACTIONS.  [SECT.  xvi. 

2.  What  is  the  value  of  ^  of  a  shilling  ?  Ans.  3£d. 

3.  What  is  the  value  of  $  of  a  guinea,  at  28  shillings  ? 

Ans.  21s.  9d.  l£qr. 

4.  What  is  the  value  of  /T  of  a  cwt.  ? 

Ans.  2qr.  151b.  4oz.  5T9Tdr. 

5.  What  is  the  value  of  f  of  a  Ib.  avoirdupois  ? 

Ans.  7oz.  l|dr. 

6.  What  is  the  value  of  f  of  a  Ib.  troy  ? 

Ans.  lOoz.  13dwt.  8gr. 

7.  What  is  the  value  of  ^  of  a  Ib.  apothecaries'  weight. 

Ans.  3 §  53  19 

8.  What  is  the  value  of  ^  of  a  yard  ? 

Ans.  2qr.  Ona.  1 

9.  What  is  the  value  of  £  of  an  ell  English  ? 

Ans.  2qr.  3na. 

10.  What  is  the  value  of  |£  of  a  mile  ? 

Ans.  6fur.  30rd.  12ft. 

11.  What  is  the  value  of  f  of  a  furlong  ? 

Ans.  35rd.  9ft.  2in. 

12.  What  is  the  value  of  fa  of  an  acre  ? 

Ans.  2R.  6rd.  4yd.  5ft.  127^in. 

13.  What  is  the  value  of  T9T  of  a  rod  ? 

Ans.  144ft.  19Tyin. 

14.  What  is  the  value  of  -^  of  a  cord  ? 

Ans.  9ft.  1462f£in. 

15.  What  is  the  value  of  •&  of  a  hhd.  of  wine  ? 

Ans.  6gal.  2qt.  Ipt.  0-^gi. 

16.  What  is  the  value  of  £  of  a  hhd.  of  beer  ? 

Ans.  42gal. 

17.  What  is  the  value  of  ££  of  a  year  (365£  days)  ? 

Ans.  174d.  16h.  26m.  5/3sec. 

18.  What  is  the  value  of  7^  of  a  dollar  ? 

Ans.f7.74AV 

CASE  XII. 

To  reduce  any  mixed  quantity  of  weights,  measures,  &c.,  to 
the  fraction  of  the  integer. 

1.  What  part  of  a  shilling  is  Id.  ?    is  2d.  ?     is  3d.  ?    is  4d.  ? 

2.  What  part  of  a  pound  is  2s.  ?    3s.  ?   4s.  ?   5s.  ?    6s.  ?   9s.  ? 

3.  What  part  of  a  furlong  is  3rd.  ?   4rd.  ?  5rd.  ?  8rd.  ?   9rd.  ? 

4.  What  part  of  a  hogshead  is  5gal.  ?    8gal.  ?    lOgal.  ? 

5.  What  part  of  a  foot  is  2  inches  ?   3in.  ?  4in.  ?  5in.  ?  8in.  ? 


SECT,  xvi.]  VULGAR  FRACTIONS.  109 

6.  What  part  of  a  £.  is  5s.  5d.  lT9Tqr.  ? 

s.       d.        qr. 

20  5     5     1-^f         In    this   question,   the   shillings, 

12  12  pence,  farthings,  &c.  are  reduced 

240  65  to  elevenths  of  farthings  for  the  nu- 

4  4  merator  of  a  fraction.     A  pound  is 

2g}  also  reduced  to   the  same  denomi- 

nation, for  a  denominator.      The 
fraction  is  then  reduced  to  its  lowest 


10560        2880  termg- 


RULE.  —  Reduce  the  given  number  to  the  lowest  denomination  it  con- 
tains for  a  numerator  ;  and  then  reduce  the  integer  to  the  same  denom- 
ination, for  the  denominator  of  the  fraction  required. 

7.  Reduce  3£d.  to  the  fraction  of  a  shilling.         Ans.  /¥. 

8.  Reduce  21s.  9d.  1^-qr.  to  the  fraction  of  a  guinea. 

Ans.  J. 

9.  Reduce  2qr.  151b.  4oz.  5T9Tdr.  to  the  fraction  of  a  cwt. 

Ans.  -j7T. 

10.  What  part  of  a  pound  are  7oz.  l£dr.  ?  Ans.  £. 

11.  What  part  of  a  pound  troy  are  lOoz.  13d  wt.  8gr.  ? 

Ans.  f  . 

12.  What  part  of  a  pound  apothecaries'  weight  are  3§  55 

r.  ? 


19  12T^gr.  ?  Ans.TV 

13.  What  part  of  a  yard  are  2qr.  Ona.  1-^in.  ?     Ans.  T7jj-. 

14.  What  part  of  an  ell  English  are  2qr.  3na.  0|in.  ? 

Ans.  f  . 

15.  What  part  of  a  mile  are  6fur.  30rd.  12ft.  Sin.  0|§br.  ? 

Ans.  H- 

16.  Reduce  35rd.  9ft.  2in.  to  the  fraction  of  a  furlong. 

Ans.  f  . 

17.  What  part  of  an  acre  are  2R.  6rd.  4yd.  5ft.  127TVn.  ? 

Ans.  A. 

18.  What  part  of  a  square  rod  are  144ft.  19^^.  ? 

Ans.  ^-. 

19.  What  part  of  a  cord  are  9ft.  1462f^in.  ?         Ans.  •&. 

20.  What  part  of  a  hogshead  of  wine  are  6gal.  2qt.   Ipt. 
0T\gi.  ?  Ans.  T\. 

21.  What  part  of  a  hhd.  of  beer  are  42gal.  ?  Ans.  f. 

22.  What  part  of  a  year  (365£  da.)  are  174d.  16h.26m. 
sec.  ?  Ans.  % 

10 


O  VULGAR  FRACTIONS.  [SECT.  XTII. 

SECTION  XVII. 
ADDITION  OF  VULGAR  FRACTIONS. 

CASE  I. 

To  add  fractions  that  have  a  common  denominator. 

1.  What  part  of  an  apple  is  £  and  £  ?  £  and  £  ?  f  and  £  ? 

2.  What  part  of  a  dollar  is  |  and  f  ?  T2^  and  &  ?  ^  and  T^  ? 

3.  What  part  of  a  shilling  is  T3j  and  -£r  ?  TV>  TV,  and  &  ? 

4.  What  part  of  an  orange  is  |  and  f  ?  £  and  £  ?  f  and  f  ? 

5.  Add  T3z,  ^,  A.  and  ri  together. 

OPERATION. 

7       ll  =       =      =  2     Ans. 


In  this  question,  we  add  the  numerators,  and  divide  their  sum 
by  the  denominator. 

RULE.  —  Write  the  sum  of  the  numerators  over  the  common  denom- 
inator, and  reduce  the  fraction  if  necessary. 

6.  Add  TV,  TV,  TV,  «i  If,  and  ff  together.  Ans.  3ff  . 

7.  Add  7V  &>  H»  it.  and  if  together.  Ans.  2JJ. 

8.  Add  ^,  |f,  |f,  |f,  and  Jf  together.  Ans.  2/T. 

9.  Add  T%,  TV9T,  1%,  and  «l  together.  Ans.  If. 

10.  Add  fH,  |Sf  ,  ^Jf  ,  and  ^T  together.  Ans.  2||f  . 

11.  Add  fff  ,  f  |f  ,  fff  ,  and  f  |f  together.  Ans.  3?f  f. 

12.  Add  |ff|,  ff«,  and  |f  «  together.  Acs. 

13.  Add  £!f|,  Jf«,  and  f  §f  J  together.  Ans. 

14.  Add  T^ftfr,  ^%,  and  f^  together.  Ans. 


CASE  II. 

To  add  fractions  that  have  not  a  common  denominator. 
1.  What  is  the  sum  of  £,  T^-,  f  i»  and  £$  ? 
First  Method. 

OPERATION. 

4)8  12  16  20     4x2x3x2x5  =  240  common  denominator. 
2)2     3     4     5  8 

1325  12 

16 
20 


30X   7  =  210 
20X   5=100 


=  165 
12X13  =  156 


Having  found  a  common  de-  631 

nominator   by  Case  VIII. ,  we  240  ~    ^A 

proceed  as  in  the  last  Case. 


Ans. 


SECT.    XVII.] 


VULGAR  FRACTIONS. 


Ill 


Second  Method. 


OPERATION. 

7  X  12  X  16  X  20  =  26880 
5x   8X  16X20=12800 

11  X   8X  12x20  —  21120 

13  X  8  X  12  X  16  =  19968 

80768 

8  X  12  X  16  X  20  =  30720 


Let  the  pupil  exam- 
ine the  second  method 
of  reducing  fractions 
to  a  common  denomi- 
nator in  Case  VIII., 
Sec.  XVI. 


RULE.  —  Reduce  mixed  numbers  to  improper  fractions,  and  compound 
fractions  to  simple  fractions  ;  then  reduce  all  the  fractions  to  a  common 
denominator,  and  the  sum  of  their  numerators  written  over  the  common 
denominator  will  be  the  answer  required. 

2.  Add  £,  £,  fa  and  |  together.  Ans.  2*-f  . 

3.  Add  T7T,  A,  I,  and  £  together.  Ans.  l|f£. 

4.  Add  T30,  T*F,  fa  and  J  together.  Ans.  2Jg$. 

5.  Add  T5T,  -jj^,  y1^-,  and  ^  together.  Ans.  1. 

6.  Add  Tf  ,  ft  ,  J$,  and  T9;  together.  Ans.  3£$. 

7.  Add  £,  £,  £,  £,  £,  and  |  together.  -  Ans.  1T8¥30. 

8.  Add  f,  f  ,  and  5f  together.  Ans.  6f  &£  . 

9.  Add  ^-,  /2-,  and  9T3T  together.  Ans.  9f  £. 

10.  Add  f  ,  f  ^  and  4^  together.  Ans.  6^. 

11.  Add  f,  7^,  and  8J  together.  Ans. 

12.  Add  J,  3|,  and  5f  together.  Ans. 

13.  Add  6f  ,  7f  ,  and  4f  together.  Ans.  18f  ££. 

NOTE.  —  If  the  quantity  be  a  mixed  number,  -the  better  way  is  to  add 
their  fractional  parts  separately,  as  in  the  following  example. 

14.  What  is  the  sum  of  llf  ,  15|,  12yV,  and  17f  ? 


4)4     8  12     6 

3)1     2     3     6 

2)1     2     1     2 

1111 


OPERATION. 

4x3x2  =  2 
llf 

Ans.  =  57^ 

4 

6  X  3  =  18 
3X7  =  21 
2  X  5  =  10 
4  X  5  =  20 
69 
24 

15.  What  is  the  sum  of  llf,  19^,  and  23f  ?  Ans. 

16.  What  is  the  sum  of  18|,  27-^,  and  49 J  ?        Ans."  96. 

17.  What  is  the  sum  of  21f,  18f,  and  26£  ?      Ans.  66T|. 


1-4  VULGAR  FRACTIONS.  [SECT.  xvn. 

18.  What  is  the  sum  of  17f,  14£,  and  13f  ?    Ans.  45T£B. 

19.  What  is  the  sum  of  16f  ,  8J,  9f  ,  3£,  and  If  ? 

Ans.  40TV 

20.  What  is  the  sum  of  371  ^,  614^§,  and  81|  ? 

Ans.  1068f£. 

21.  Add  f  of  18T3r,  and  j£  of  f  of  6T3T  together. 

Ans.  12ff-}. 

22.  Add  |  of  18,  and  ^  of  J4-  of  7^  together. 

Ans.  13%±. 

23.  Add  4  of  15±,  and  f  of  107f  together.          Ans.  93f 


24.  Add  £  of  f  of  28       to  a.  Ans.  6T9^. 

25.  Add*,2|f  --,  and  together.  Ans.  8jH|H»- 


CASE  III. 

To  add  any  two  fractions,  whose  numerators  are  a  unit. 
RULE.  —  Place  the  sum  of  the  denominators  over  their  product. 
EXAMPLE. 

1.  Additof  4  +  5=  9 


NOTK.  —  The  truth  of  this  rule  is  evident  from  the  fact,  that  this  pro- 
cess reduces  the  fractions  to  a  common  denominator,  and  then  adds  the 
numerators. 

If  the  numerators  of  the  given  fractions  be  alike,  and  more 
than  a  unit,  multiply  the  sum  of  the  denominators  by  one  of  the 
numerators  for  a  new  numerator,  then  multiply  the  denomina- 
tors together  for  a  new  denominator. 


SECT,  xvn.]  VULGAR  FRACTION 


10.  Add  f  to  f  .  4  +  5  =  9  X3 

4  X  5 

11.  Add  fctoM  to  M  to  A,f  to  f,  f  to  f,  f  to  A,  ^  to  f 

12.  Add  f  to  T3r,  f  to  f  ,  I  to  f  ,  f  to  |,  f  to  f  ,  |  to  A- 

13.  Add  f  tO  f  ,  f  tO  T6r,  f  tO  -ft,  T6T  tO  Ai  A  tO  A»  T6T  tO  A- 

14.  Add  f  tO  A,  T8T  tO  T8T.  TT  tO  A,  A  tO  T85  .  A  to  T8T,  A  to  A- 

15.  Add  A  tO  T9T,  A  ^  T9*,  T9*  tO  A,  T9T7  to  T9T,  A  *  A- 

16.  Add  I-  to  I  ,  ^  to  A>  I  to  T7T,  I-  to  A»  I-  to  A>  f  to  A- 

17.  Add  f  to  Ai  f  to  A.  f  to  A>  t  to  A,  f  to  A»  t  to  A- 

18.  Add  «  to  ^,  if  to  «,  «  to  it,  if  to  ^,  it  to  if 

19.  Add  i£  to  ii,  i£  to  ii,  «.  to  ii,  «  to  ii,  «  to  i|. 

NOTE.  —  The  preceding  rule  may  be  found  very  useful,  because  all 
similar  questions  may  be  readily  performed  mentally. 

CASE  IV. 

To  add  compound  numbers. 

1.  Add  A  of  a  £.  to  T9T  of  a  £. 

Value  of  A  of  a  «£.  ==  10s.  9d.  Oi§qr.  This  question  is  per- 
Value  of  T9T  of  a  £.=  16s.  4d.  lT5Tqr.  formed  by  finding  the 

1£.     7s.  Id.  2T5Aqr.  values  of  A  of  a  £- 
and  A  of  a  £.  by  Case 

XL,  Sec.  XVI.  The  fractions  if  and  T5T  are  added  by  Case  II. 
of  Addition  of  Fractions.  The  following  questions  are  per- 
formed in  the  same  manner. 

The  above  question  may  be  performed  by  first  adding  the 
fractions  of  the  pounds  together,  and  then  finding  their  value  by 
Case  XL  ;  thus  : 

OPERATION. 

A*-  +  A&  =  if^-  =  !<£•  7s-  ld-  *&s<F'  Ans- 

2.  Add  together  |  of  a  £.,  f  of  a  <£.,  and  f  of  a  shilling. 

OPERATION. 

£.      s.      d.        qr. 

f  of  a  £.  =  0  8  10  2f 
fofa«£.  —  0  8  6  3f 
|  of,  as.  _  0  0  4  3J 


0  17  10 

The  above  question  may  be  solved  by  first  reducing  f  of  a 
shilling  to  the  fraction  of  a  pound  by  Case    IX.,    Sec.     XVI. 
10* 


114  VULGAR  FRACTIONS.  [SECT.  XYIII. 

and  then  adding  it  to  the  other  numbers,  and  finding  their  value 
by  Case  XL,  Sec.  XVI.     Thus  : 

f  of  a  shilling  =  f  X  &  =  ?$?  =  Ar& 
|  £.  _J-  f  ,£._{_  ^  £.  =  §*tf£.  =  0£.  17s.  lOd.  l-A^qr.  Ans. 


NOTE.  —  The  pupil  should  solve  the  following  questions  by  both  pro- 


3.  Add  together  -^  of  a  ton,  and  £J  of  a  cwt. 

Ans.  13cwt.  2qr. 

4.  Add  together  f  of  a  yard,  f  of  an  ell  English,  and  ^  of 
a  qr.  Ans.  3qr.  3na.  l^f§in. 

5.  Add  together  -fa  of  a  mile,  ^  of  a  furlong,  and  ^  of  a 
yard.  Ans.  5fur.  16rd.  Oft.  3in.  Uffbar. 

6.  A.  has  three  house-lots ;  the  first  contains  £  of  an  acre, 
the  second  §  of  an  acre,  and  the  third  -f£  of  an  acre.     How 
many  acres  do  they  all  contain  ? 

Ans.  2A.  1R.  9p.  142ft?  87£in. 

7.  A  man  travelled  18£  miles  the  first  day,  23|±  miles  the 
second  day,  and  lO^  miles  the  third  day.     How  far  did  he 
travel  in  the  three  days  ?         Ans.  61m.  2fur.  3rd.  13ft.  4f  in. 

8.  Add  -H-  of  a  gallon  of  wine  to  -^  of  a  hhd. 

Ans.  6gal.  Oqt.  Ipt.  l£gu 

9.  Add  -j^  of  a  week  to  £  of  a  day.         Ans.  2d.  9h.  iMn. 

10.  Add  f  of  a  square  foot  to  £  a  foot  square.      Ans.  1  foot. 

11.  Add  6  inches  to  llrd.  16ft.  5in.        Ans.  12rd.  Oft.  5in. 


SECTION  XVIII. 

SUBTRACTION  OF  VULGAR  FRACTIONS. 
CASE  I. 

To  subtract  fractions  that  have  a  common  denominator. 

1.  If  £  be  taken  from  f  what  will  be  left  ? 

2.  If  |  be  taken  from  |  what  will  be  left  ? 

3.  If  -fa  be  taken  from  ^  what  will  be  left  ? 

4.  What  portion   of  a  dollar   will   be   left,  if  ±  be   taken 
from  £  ? 

5.  Subtract  -fa  from  -}•£. 

11  —  5  =  6.          = 


SECT.  XYIH.]                   VULGAR  FRACTIONS.  115 

RULE.  —  Subtract  the  less  numerator  from  the  greater,  and  under  the 
remainder  write  the  common  denominator,  and  reduce  the  fraction  if 
necessary. 

6.  Subtract  T6T  from  if.  Ans.  yV 

7.  Subtract  -fa  from  if.  Ans.  T9?. 

8.  Subtract  -^  from  ff .  Ans.  if. 

9.  Subtract  £J  from  if.  Ans.  /F. 

10.  Subtract  if  from  f  f .  Ans.  ^. 

11.  Subtract  if  from  |f.  Ans.  if. 

12.  Subtract  if  from  ff .  Ans.  if-. 

13.  Subtract  iif  from  iif.  Ans.  TfT. 

14.  Subtract  T^  from  f^j.  Ans.  £. 

15.  Subtract  if-  from  ff .  Ans.  £. 

16.  Subtract  yW  from"T3¥6¥.  Ans.  -fa. 

17.  Subtract  y1^  from  y*^.  Ans.  fa. 

18.  Subtract  y3^  from  T5T9A.  Ans. 

19.  Subtract  ,4^  from  ^AV  Ans. 


CASE  II. 
To  subtract  fractions  whose  denominators  are  unlike. 

1.  Subtract  f  from  if. 

OPERATION. 

Common  denominator  77 In   this  ques- 

11    7  x  10  =  70  ti°n'  we  ^n(^  ^e 

\\  y^    4  =  44  common  denom- 

2g  inator,    77,    by 

—  =  Ans.    multiplying   the 
77  two     denomina- 

tors, 7  and  11 ;  and  then  obtain  the  numerators,  as  in  Case  VIII., 
Sec.  XVI. ;  the  difference  of  which  we  write  over  the  common 
denominator. 

2.  From  T£  take  yV  Ans.  if. 

OPERATION. 

'  — — —    4  X  4  X  3  =  48  common  denominator. 

16 
12 


4     3 


3x11  —  33 
4x    5  =  20 
13 


48~AnS- 
3.  From  9^  take  5T£.  Ans. 


116 


VULGAR  FRACTIONS. 

OPERATION. 


[SECT,  xviii. 


8)8    16 

1    2 


8  X  2  =.  16  common  denominator. 


8 
16 


2  x  79  =  158 
1  x91  =    91 


4.  From  J  of  16£  take  T5Z  o 

OPERATION. 


Ans.  2££. 


RULE.  —  Reduce  compound  fractions  to  simple  ones,  and  mixed  ones 
to  improper  fractions ;  then,  having  found  a  common  denominator,  di- 
vide this  by  each  of  the  denominators  of  the  fraction,  and  multiply  their 
quotients  by  their  respective  numerators.  The  difference  of  these  prod- 
ucts, placed  over  the  common  denominator  t  will  give  the  answer  required. 

Ans.  «. 

Ans.  f  J. 
Ans.  \. 

Ans.  ££. 

Aits,  fa 
Ans.  £f£. 
Ans.  $§1. 
Ans.  ^r. 

Ans.  /$. 
Ans.  ^V 

Ans.  •&. 

Ans.  -5^-. 
Ans.  l^9ff. 
Ans.  1|^. 

Ans. 


5.  From  •/,-  take  f . 

6.  From  f  "take  f . 

7.  From  £  take  £. 

8.  From  -f  f  take  ^-. 

9.  From  ^  take  ^-. 

10.  From  |f  take  T^. 

11.  From  jf  take  &. 

12.  From  ^  f  take  ^T. 

13.  From  |^  take  A- 

14.  From  ££  take  ^. 

15.  From  ^  take  3%-. 

16.  From  A  take  T3F. 

17.  From  7J  take  ^  of  9. 

18.  From  f  of  8f  take  f  of  5. 

19.  From  |  of  3  take  £  of  2. 


SECT.  XVIH.]  VULGAR  FRACTIONS.  117 

CASE  III. 

To  subtract  a  proper  fraction  or  a  mixed  one  from  a  whole 
number. 

1.  From  $  7  take  $  3$. 

OPERATION.      To  subtract  f  in  this  example  we  must  borrow  1 
7  from  the  7  in  the  minuend,  and  reduce  it  to  eighths 

3f  (f  )  and  the  f  must  be  taken  from  them  ;  |-  from  f 

Ans.  leaves  f  •  To  pay  for  the  1  which  was  borrowed, 
1  must  be  added  to  the  3  in  the  subtrahend,  1  4-  3 
=  4,  and  4  taken  from  7  leaves  3,  and  the  f  placed  at  the  right 
hand  of  it  gives  the  answer  $  3f  . 

By  adopting  the  following  rule,  the  same  result  will  be  ob- 
tained. 

RULE.  —  Subtract  the  numerator  from  the  denominator  of  the  frac- 
tion, and  under  the  remainder  write  the  denominator,  and  carry  one  to 
the  whole  number  of  the  subtrahend  to  be  subtracted  from  the  minuend. 

OPERATION. 
2.  3.  4.  5.  6.  7. 

32  16          671  385  16  18 

*6«  °Hf       Jf 

26f         11£        670-&        368ff 


8.  9.  10.  11.  12.  13. 

19          27  169  711  46          81 


If  it  be  required  to  subtract  one  mixed  number  from  another 
mixed  number,  the  following  method  may  be  adopted. 

14.  From  8f  take  4f  .  Ans.  3|f  . 

OPERATION.  In  this  question,  we  multiply  the  3  and  the  7,  the 
8f  =  8£f-  numerator  and  the  denominator  of  the  fraction  in 
4f-  =  4f  |  the  minuend,  by  5,  the  denominator  of  the  fraction 
~3p;  in  the  subtrahend,  and  we  have  a  new  fraction  -£f  , 
which  we  write  at  the  right  hand  of  the  8,  thus, 
8^|..  We  then  multiply  the  numerator  and  denominator  of  the 
subtrahend  by  7,  the  denominator  of  the  minuend  ;  and  we  have 
another  new  fraction,  f  f  ,  which  we  place  at  the  right  hand  of 
the  4,  thus,  4§f.  It  will  now  be  perceived,  that  we  have 
changed  the  fractions  8f  and  4f  to  other  fractions  of  the  same 
value,  having  a  common  denominator.  We  now  subtract  as  in 
question  1,  by  adding  1  (=  $f)  to  ^f,  which  makes  £$,  and 


118  VULGAR  FRACTIONS.  [SECT.  xvin. 

from  this  we  subtract  ff  ;  thus,  £f  —  ff  =  ff  .  We  then 
carry  the  1  we  borrowed  to  the  4,  1  -j-  4  =  5,  which  we  take 
from  8,  and  find  3  remaining.  The  answer,  then,  is  3-ff  . 

If  the  fraction  in  the  subtrahend  be  less  than  the  fraction  in 
the  minuend,  we  proceed  as  in  the  following  problem. 

15.  From  9f  take  3^-.  Ans.  6£|. 

OPERATION.          Having  reduced  the  fractions  to  a  common  de- 
9£    —  9££     nominator,  as  in  the  last  problem,  we  subtract  35, 
3^.  —  3f  |-     the  numerator  of  the  subtrahend,  from  48,  the  nu- 
gis     merator  of  the  minuend,  and  the  remainder  13  we 
write  over  the  common  denominator  =  ^-$,  which 
we  annex  to  the  difference  between  9  and  3,  =  6  ;  thus,  6££  . 

16.  17.  18.  19.  20.  21. 

cwt.  Tuns.  8  lb.  oz.  Miles. 

From  18f    73£    67£    29|f    144£    171  £4 
Take  9f    16|£   16f 


22.  23.  24.  25.  26.  27. 

Furlongs.  Rods.          Inches.          Feet.  Bushels.  Pecks. 

From  101£  165£ 

Take    93£  98£ 


28.  From  a  hhd.  of  wine  there  leaked  out  7T9T  gallons  ;  what 
quantity  remained  ?  Ans.  55T2Tgal. 

29.  A  man  engaged  to  labor  30  days,  but  was  absent  5/j 
days  ;  how  many  days  did  he  work  ?  Ans.  24^da. 

30.  From  144  pounds  of  sugar  there  were  taken  at  one  time 
17|  pounds,  and  at  another  28/j  pounds  ;  what  quantity  re- 
mains ?  Ans.  97£f  lb. 

31.  A  man  sells  9£  yards  from  a  piece  of  cloth  containing 
34  yards  ;  how  many  yards  remain  ?  Ans.  24£yd. 

32.  The  distance  from  Boston  to  Providence  is  40  miles.     A. 
having  set  out  from  Boston,  has  travelled  ^-  of  the  distance  ; 
and  B.  having  set  out  at  the  same  time   from  Providence,  has 
gone  -/Y  of  the  distance  ;  how  far  is  A.  from  B.  ? 

Ans.  28T|Tm. 

33.  From  £  of  a  square  yard  take  £  of  a  yard  squared. 

Ans.  2  square  feet. 

34.  What  is  the  difference  between  —  $  and  —   :-  ? 


Ans. 


SECT,  xvin.]  VULGAR  FRACTIONS.  119 

CASE  IV. 

To  subtract  one  fraction  from  another,  when  both  fractions 
have  a  unit  for  a  numerator. 

OPERATION. 

I.  Take  4-  from  4.  ^^5  ==  Jt   Ans. 

7x3       ** 

The  student  will  perceive,  that  this  operation  reduces  the 
fractions  to  a  common  denominator. 

RULE.  —  Write  the  difference  of  the  denominators  over  their  product. 

2.  Take  £  from  £,  £,  £,  |,  £,  f  ,  £  ;  2V  from  TV,  TV,  TV>  A- 

3.  Take  £  from  £,  £,  £,  £,  |,  £,  |  ;  TV  from  TV,  TV,  iV 

4.  Take  |  from  |,  £,£,£;  TV  from  £,  £,  1,  f  ,  |. 

5.  Take  |  from  £,  £,  £  ;  }  from  J,  £,  J,  |. 

6.  Take  T»T  from  TV,  |,  -£,  i,  |,  i,  f,  i,  ^. 

7.  Take  ^  from  £  ;  £  from  £,  -J  ;  ^  from  £. 

8.  Take  TV  from  £,  ^,  ^,  ^,  ^,  |,  1,  i,  ^,  1V. 

9.  Take  TV  from  &  $,  ^  |,  ^  },  £,  ^  ^,  ^ 
10.  Take  TV  from  ^,  ^,  £,  -i,  J,  |,  4,  £. 

NOTE.  —  If  the  numerators  of  the  given  fractions  be  alike,  and  more 
than  a  unit,  multiply  the  difference  of  the  denominators  by  one  of  the  nu- 
merators for  a  new  numerator,  then  multiply  the  denominators  together 
for  a  new  denominator. 

OPERATION. 

II.  Take  f  from  f.  7  —  3  =  4;4X2_ 


12.  Take  f  from  f  ;  f  from  f-  ;  -&  from  f  ;  -&  from  f  . 

13.  Take  f  from  f  ;  T4T  from  f  ;  T4T  from  f  ;  T*T  from  TV 

14.  Take  %  from  £  ;  T5T  from  £  ;  T5T  from  f  ;  ^\-  from  -fa- 

15.  Take  T6T  from  f  ;  T6T  from  f  ;  T67  from  f-  ;  •&  from  T6T. 

16.  Take  T\  from  ^  ;  -^  from  f  ;  T^  from  f  ;  T4^  from  T4T. 

17.  Take  T8T  from  |  ;  T8T  from  f  ;  -^  from  f  ;  T8T  from  f  . 

18.  Take  fT  from  f  ;  T2T  from  f  ;  ^-  from  f  ;  T2T  from  f  . 

19.  Take  |$  from  ^  ;  |f  from  ^  ;  ^  £  from  if  ;  |£  from  tf. 

20.  Take  -^  from  J£  ;  V2-  ^om  JT2-  ;  J^  from  -'/  ;  J-§  from  4^-. 

NOTE.  —  The  above  questions,  and  those  of  a  similar  kind,  may  readily 
be  performed  mentally. 


120  VULGAR   FRACTIONS.  [SECT.  xvin. 

CASE  V. 

To  subtract  compound  numbers. 
1.  From  /y  of  a  £.  take  f  of  a  £. 

OPERATION. 

33  common  denominator 

24 

To  perform  this  question, 


Value  of  A  £.  =  12s. 
Value  of  f  «£.  =    4s.  5  £  d. 
Ans.    8s. 


TO  we  find  by  Case  XI.,  Sect. 
i|  XVI.,the  value  of ^£.=  12s. 
33  8y8Td. ;  and  also  of  f  £.  =  4s. 
5^d. ;  we  then    find  a  com- 
mon denominator  of  the  fractional  part,  by  multiplying  together 
their  denominators,    1 1  X  3  —  33.      We  then  proceed  as   in 
Case  II.,  Sect  XVIII. 

This  question  can  be  performed  by  first  subtracting  the  frac- 
tion f  of  a  <£.  from  -/y  of  a  £.,  and  then  reducing  the  remainder 
by  Case  XL,  Sect.  XVI. ;  thus  : 

•/y  £.  — f  £.  =  (£^£.  =  0<£.  8s.  3d.  l£§qr.  Ans. 
2.  From  7^-  of  a  ton  take  £  of  a  cwt. 
OPERATION. 

cwt.    qr.    Ib.    oz.       dr. 

•&  of  a  ton  =  1     1     4    8 
4  of  a  cwt.  =        3599 


Ans.  0     1  26  14 
This  question  may  also  be  performed  by  first  reducing  £  of  a 
cwt.  to  the  fraction  of  a  ton  by  Case  IX.,  Sect.  XVI.,  and  sub- 
tracting it  from  TJ\  of  a  ton,  and  then  reducing  the  remainder  to 
its  proper  terms  by  Case  XI.,  Sect.  XVI.     Thus  : 


^  —  2^  =  ^  of  a  ton  =  Iqr.  261b.  14oz.  lO^dr.  Ans. 

RULE.  —  Find  the  value  of  the  fractions  in  integers  ;  then  subtract 
as  in  the  foregoing  rules. 

3.  From  £  of  an  ell  English  take  f  of  a  yard. 

Ans.  3qr.  Ona.  2^in. 

4.  From  f  of  a  mile  take  -/y  of  a  furlong. 

Ans.  Ifur.  5rd.  10ft.  lOin. 

5.  From  f  of  a  degree  take  f  of  a  mile. 

Ans.  49m.  Ofur.  13rd.  lift.  9in.  l^bar. 


SECT,  xiii.]  DIVISION.  73 

The  rule  for  pointing  off  cents  and  mills  is  the  same  as  in  Multiplication. 

If  the  dividend  consist  of  dollars  only,  and  be  either  smaller  than  the 
divisor,  or  not  divisible  by  it  without  a  remainder,  annex  two  or  three 
ciphers,  as  the  case  may  require,  and  the  quotient  will  be  cents  or  mills 
accordingly. 

1.  If  97  bushels  of  wheat  cost  $  147.82,8,  what  is  the  value 
of  one  bushel  ?  Ans.  $  1.52,4. 

OPERATION. 

97)  147.82,8($  1.52,4 
97_ 
508 

485 

232 
194 

388 
388 

2.  Bought  1789  acres  of  land  for  $  1699.55  ;  what  cost  one 
acre?  Ans.  $0.95. 

3.  A  trader  sold  425  pounds  of  sugar  for  $  51.00  ;  what  was 
the  cost  of  one  pound  ?  Ans.  $  0.12. 

4.  When  rye  is  sold  at  the  rate  of  628  bushels  for  $  471.00, 
what  is  that  a  bushel  ?  Ans.  $  0.75. 

5.  A  merchant  bought  329  yards  of  broadcloth  for  $  904.75  ; 
what  cost  one  yard  ?  Ans.  $  2.75. 

6.  When  a  chest  of  tea  containing  42  pounds  can  be  bought 
for  $  31.50,  what  cost  one  pound  ?  Ans.  $  0.75. 

7.  If  it  cost  $1460  to    support  a  family  365  days,   what 
would  be  the  expense  per  day  ?  Ans.  $  4.00. 

8.  A  shoe-dealer  sold  125  cases  of  shoes  for  $  2500  ;  what 
was  the  cost  per  case  ?  Ans.  $  20.00. 

9.  A  flour-merchant  sold  475  barrels  of  flour  for  $2018.75; 
what  cost  one  barrel  ?  Ans.  $  4.25. 

10.  Bought  42  barrels  of  pears  for  $  73.50  ;  what  cost  one 
barrel  ?  Ans.  $  1.75. 

11.  If  1624  pounds  of   pork  cost  $97.44,  what  cost  one 
pound  ?  Ans.  $  0.06. 

12.  If  47000  shingles  cost  $  176.25,  what  is  the  cost  per 
thousand  ?  Ans.  $  3.75. 

13.  Bought  148  tons  of  plaster  of  Paris  for  $  337.44  ;  what 
was  it  per  ton  ?  Ans.  $  2.28. 

14.  If  78  barrels  of  fish  cost  $  303.42,  what  will  one  barrel 
cost  ?  Ans.  $  3.89. 

7 


74  UNITED  STATES  MONEY.  [SECT.  xm. 

15.  A  farmer  sold  691  bushels  of  wheat  for  $  863.75 ;  what 
was  it  per  bushel  ?  Ans.  $  1.25. 

16.  If  a  man  earn  $  434.35  in  a  year,  what  is  that  per  day? 

Ans.  $  1.19. 

17.  Sold  169  tons  of  timber  for  $  790.92 ;  what  cost  one 
ton  ?  Ans.  $  4.68. 

18.  What  cost  one  pound  of  leather,  if  789  pounds   cost 
8142.02?  Ans.  $0.18. 

19.  If  369  tons  of  potash  cost  $48910.95,  what  will  be  the 
price  of  one  ton  ?  Ans.  $  132.55. 

20.  Bought  47  hogsheads  of  salt,  each  hogshead  containing 
7  bushels,  for  $  368.48  ;  what  cost  one  bushel  ?    Ans.  $  1.12. 

21.  If  19  cords  of  wood  cost  $  106.97,  what  cost  one  cord  ? 

Ans.  $5.63. 

22.  When  19  bushels  of  salt  can  be  bought  for  $  30.87,5, 
what  cost  one  bushel  ?  Ans.  $  1.62,5. 

23.  If  17  chests  of  souchong  tea,  each  weighing  59  pounds, 
cost  $ 672.01,  what  cost  one  pound?  Ans.  $ 0.67. 

24.  Sold  73  tons  of  timber  for  $  414.64 ;  what  did  I  receive 
per  ton  ?  Ans.  $  5.68. 

25.  Bought  oil  at  the  rate  of  144  gallons  for  $  234.00 ;  what 
did  I  give  per  gallon  ?  Ans.  $1.62,5. 

26.  A  landholder  sold  47  acres  of  land  for  $  1774.25 ;  what 
did  he  receive  per  acre  ?  Ans.  $  37.75 

27.  What  is  the  price  of  one  yard  of  broadcloth,  if  163  yards 
cost  $1106.77?  Ans.  $6.79. 

28.  If  a  farm,  containing  144  acres,  is  valued  at  $  10043.71,2, 
what  is  one  acre  worth  ?  Ans.  $  69.74,8. 

BILLS. 

1.  Boston,  July  4,  1835. 
Mr.  James  Dow, 

Bought  of  Dennis  Sharp, 

17  yds.  Flannel,  at  .45  cts. 

19    "    Shalloon,  "  .37    " 

16  "    Blue  Camlet,  "  .46    " 
13    "    Silk  Vesting,  "  .87    " 

9  "  Cambric  Muslin,  "  .63  " 

25  "  Bombazine,  "  .56  " 

17  "  Ticking,  "  .31  " 
19  "  Striped  Jean,  "  .16  " 

$61.33 
Received  payment, 

Dennis  Sharp. 


SECT.    XIII.] 

2. 

Mr.  Samuel  Smith, 


BILLS. 


13  Ibs.  Tea, 

16  "  Coffee, 

36  "  Sugar, 

47  "  Cheese, 

12  "  Pepper, 
7  "  Ginger, 

13  "  Chocolate, 


75 


Haverhill,  May  5,  1835. 


Bought  of  David  Johnson, 


at 


Received  payment,        Dayid 


.98  cts. 

.15  " 

.13  " 

.09  « 

.19  " 

.17  " 

.61  " 


$35.45. 


3. 

Mr.  John  Dow, 


17  yds.  Broadcloth, 


Salem,  February  29,  1835. 
Bought  of  Richard  Fuller, 


29 
60 
49 
18 
27 
75 
36 
49 


Cassimere, 
Bleached  Shirting, 
Ticking, 
Blue  Cloth, 
Habit  do. 
Flannel, 
Plaid  Prints, 
Brown  Sheeting, 


Received  payment, 


$5.25 
1.62 

.17 

.27 
3.19 
2.75 

.61 

.75 

.18 

8372.90. 

Richard  Fuller. 


4. 

Mr.  John  Rilley, 


10  pair  Boots, 
19     "    Shoes, 
83    "    Hose, 
47  Ibs.  Ginger, 
91    "     Chocolate, 
47   "     Pepper, 
68   "     Flour, 
27  pair  Gloves, 


at 

H 


Received  payment, 


Baltimore,  January  20,  1835. 

Bought  of  James  Somes, 
$2.75 
1.25 
1.29 
.17 
.39 
.23 
.13 
1.39 

ft  258.98. 


James  Somes. 


76  UNITED  STATES  MONEY.  [SECT.  xm. 

5.  Philadelphia,  June  11,  1835. 
Mr.  Moses  Thomas, 

Bought  of  Luke  Dow, 

27  National  Spelling-Books,  '      at     $0.19 
25  Parker's  Composition,  .27 

17  National  Arithmetics,  "  .75 
9  Greek  Lexicons,                                 3.75 
8  Ainsworth's  Dictionaries,        "         4.50 

27  Greek  Readers,  "  2.25 

18  Folio  Bibles,  "  9.87 
75  Leverett's  Caesar,  "  .31 
67  Fisk's  Greek  Grammar,  .75 
15  Folsom's  Cicero's  Orations,  "  1.12 

"¥423.09. 

Received  payment, 

Luke  Dow,  by 

Timothy  True. 

6.  Boston,  June  26,  1835. 
Dr.  Enoch  Cross, 

Bought  of  Maynard  &  Noyes, 
14  oz.  Ipecacuanha,  at     $  0.67 

23  "    Laudanum,  "          .89 

17  "    Emetic  Tartar,  "        1.25 

25  "    Cantharides,  "        2.17 

27  "    Gum  Mastic,  "          .61 

56  "    Gum  Camphor,  " 

~~$  136.94. 
Received  payment, 

Maynard  &  Noyes, 

by  Timothy  Jones. 

7.  Newburyport,  June  5,  1835. 
Mr.  John  Somes, 

Bought  of  Samuel  Gridley, 

7£  yds.  Broadcloth,  at     $  4.50 

16flbs.    Coffee,  "  .16 

18J  "      Candles,  "  .25 

30     "      Soap,  "  .17 

3     «      Pepper,  "          .19 

7J  "      Ginger,  "          .18 

$  48.01  J. 
Received  payment, 

Samuel  Gridley. 


SECT.  XIII.] 


BILLS. 


77 


8. 
Mr.  Benjamin  Treat, 


37  Chests  Green  Tea, 

41      "      Black    do. 

40      "      Chests  of  Imperial  Tea, 

13  Crates  Liverpool  Ware, 

Received  payment, 


Boston,  May  1,  1835. 

Bought  of  John  True, 
at     $25.50 
«         16.17 
97.75 
169.37 


$7718.28. 
John  True. 


9. 

Mr.  John  Cummings, 


97  bbl.  Genesee  Flour, 

167    "    Philadelphia  do. 

87    "    Baltimore      do. 

196  "   Richmond     do. 
275    "    Howard  St.  do. 

69  bu.  Rye, 
136  "    Virginia  Corn, 

68  "    North  River  do. 
169  "    Wheat, 

76TonLehighCoal, 

89    "    Iron, 

49  Grindstones, 

39  Pitchforks, 

197  Rakes, 
86  Hoes, 
78  Shovels, 

187  Spades, 
91  Ploughs, 
83  Harrows, 
47  Handsaws, 
35  Millsaws, 
47  cwt.  Steel, 
57    "   Lead, 

Received  payment, 
7* 


New  York,  July  11,  1835. 
Bought  of  Lord  &  Secomb. 


at 


$6.25 
5.95 
6.07 
5.75 
7.25 
1.16 
.67 
.76 
1.37 
9.67 
69.70 
3.47 
1.61 
.17 
.69 
1.17 
.85 
11.61 
17.15 
3.16 
18.15 
9.47 
6.83 

$  17315.32. 


Lord  &  Secomb. 


78  COMPOUND  MULTIPLICATION.  [SECT.  xiv. 

SECTION  XIV. 
COMPOUND   MULTIPLICATION. 

CASE  I. 

COMPOUND  MULTIPLICATION  consists  in  multiplying  numbers 
of  different  denominations  by  simple  numbers. 

1.  What  will  6  bales  of  cloth  cost,  at  7£.  12s.  7d.  per  bale  ? 

£      8.      a.         In  this  question,  we  multiply  7d.  by  6,  and  find 

7     12    7     me  product  to  be  42d.     This  we  divide  by  12,  the 

Q     number  of  pence  in  a  shilling,  and  find  it  contains 

jc — TpT — ~     3s.  and  6d.     We  write  the  6d.  under  the  pence, 

and  cany  3  to  the  product  of  6  times  12,  and  find 

the  amount  to  be  75s.,  which  we  reduce  to  pounds  by  dividing 

them  by  20,  and  find  them  to  be  3£.  15s.     We  write  down  the 

shillings  under  the  shillings,  and  carry  3  to  the  product  of  6 

times  7«£. ;  and  we  thus  find  the  answer  to  be  45<£.  15s.  6d. 

From  the  above  illustration  we  deduce  the  following 

RULE. 

When  the  multiplier  is  less  than  12,  multiply  by  the  multiplier  and 
carry  as  in  Compound  Addition. 


2.  What  cost  9yds.  of  cloth,  at  l£.  3s.  8d.  per  yard  ? 

Ans.   10£.   13s.  Od. 

3.  What  cost  7bbls.  of  flour,  at  !<£.  8s.  7£d.  per  barrel  ? 

4.  What  cost  81bs.  of  Cayenne  pepper,  at  7s.  9£d.  per  Ib.  ? 

5.  Multiply  10yd.  3qr.  3na.  by  5. 

6.  Multiply  3cwt.  Iqr.  81b.  by  9. 

7.  Multiply  7T.  llcwt.  Iqr.  201b.  by  5. 

8.  Multiply  7  days  15h.  35m.  18sec.  by  10. 

9.  Multiply  IS£.  16s.  7£d.  by  4.  Ans.  75^.  6s.  6d. 

10.  Multiply  15£.  11s.  8|d.  by  8.     Ans.  124.£.  13s.  lOd. 

11.  Multiply  27  £.  19s.  ll£d.  by  9.     Ans.  251,£.  19s. 


12.  Multiply  19<£.  5s.  7£d.  by  11.        Ans.  212<£.  Is.  7|d. 

13.  Multiply  Sl£.  14s.  9d.  by  8.         Ans.  653<£.  18s.  Od. 

14.  Multiply  15£.  18s.  5d.  by  7.         Ans.  111£.  8s.  lid. 

15.  Multiply  13£.  5s.  4f  d.  by  12.          Ans.  159^.  4s.  9d. 

16.  Multiply  171b.  7oz.  13dwt.  13gr.  by  9. 

17.  Multiply  151b.  lloz.  19dwt.  15gr.  by  7. 


BECT.  xiv.]  COMPOUND  MULTIPLICATION.  79 

IS.  Multiply  16T.  12cwt.  3qr.  131b.  12oz.  by  11. 

19.  Multiply  13T.  3cwt.  Iqr.  141b.  13oz.  by  S. 

20.  Multiply  21b  51  53  19   16£gr.  by  8. 

21.  Multiply  47yd.  3qr.  2na.  2in.  by  7. 

22.  Multiply  17m.  7fur.  36rd.  13ft.  7in.  by  12. 

23.  Multiply  16deg.  39m.  3fur.  39rd.  5yd.  2ft.  by  9. 

24.  Multiply  16deg.  20m.  7fur.  12rd.  8ft.  llin.  l£bar.  by  6. 

25.  Multiply  16A.  2R.  4p.  19yd.  7ft.  79in.  by  11. 

26.  Multiply  7  cords  1 16ft.  1629m.  by  4. 

27.  Multiply  29hhd.  61  gal.  3qt.  Ipt.  3gi.  by  7. 

28.  Multiply  3  tuns  3hhd.  56gal.  2qt.  by  9. 

29.  Multiply  7hhd.  5gal.  2qt.  Ipt.  by  8. 

30.  Multiply  19bu.  2pk.  7qt.  Ipt.  by  6. 

31.  Multiply  36ch.  18bu.  3pk.  7qt.  by  7. 

32.  Multiply  13y.  316d.  15h.  27m.  39sec.  by  8. 

33.  If  a  man  gives  each  of  his  9  sons  23A.  3R.  19£p.,  what 
do  they  all  receive  ? 

34.  If  12  men  perform  a  piece  of  labor  in  7h.  24m.  30sec., 
how  long  would  it  take  1  man  to  perform  the  same  task  ? 

35.  If  1  bag  contain  3bu.  2pk.  4qt.,  what  quantity  do  8  bags 
contain  ? 

CASE  II. 

When  the  multiplier  is  more  than  12,  and  is  a  composite  num- 
ber, that  is,  a  number  which  is  the  product  of  two  or  more 
numbers,  the  question  is  performed  as  in  the  following 

EXAMPLE. 

36.  What  will  42  yards  of  cloth  cost,  at  6s.  9d.  a  yard  ? 

£.  a.  d.  In  this  example,  we  find  that  6 

069  multiplied  by  7  will  produce  the 

6  quantity  42  yards.  We  therefore 

206  =  price  of  6  yds.  multiply  6s  9d.  first  by  the  6,  and 

•y  then  its  product  by  7 ;  and  the  last 

— — - — ^          .       f          .     product,  14=£.  3s.  6d.  is  the  answer 
=  price  of  42  yds.  or  price  of  the  42  yards> 

The  pupil  will  now  see  the  propriety  of  the  following 
RULE. 

Multiply  by  one  of  the  factors  of  the  composite  number,  and  tlie  prod- 
uct thus  obtained  by  the  other. 

37.  What  will  16  yards  of  velvet  cost,  at  3s.  8d.  per  yard  ? 


80  COMPOUND  MULTIPLICATION.  [SECT.  XIT. 

38.  What  will  72  yards  of  broadcloth  cost,  at  19s.  lid.  per 
yard  ? 

39.  What  will  84  yards  of  cotton  cost,  at  Is.  lid.  per  yard  ? 

40.  Bought  90  hogsheads  of  sugar,  each  weighing  12cwt. 
2qr.  1  lib.  ;  what  was  the  weight  of  the  whole  ? 

41.  What  cost  18  sheep,  at  5s.  9^-d.  a  piece  ? 

42.  What  cost  21  yards  of  cloth,  at  9s.  lid.  per  yard  ? 

43.  What  cost  22  hats,  at  11s.  6d.  each  ? 

44.  If  1  share  in  a  certain  stock  be  valued  at  13£.  8s.  9£d., 
what  is  the  value  of  96  shares  ? 

45.  If  1  spoon  weigh  3oz.  5dwt  15gr.,  what  is  the  weight  of 
120  spoons  ? 

46.  If  a  man  travel  24m.  7fur.  4rd.  in  1  day,  how  far  will  he 
go  in  1  month  ? 

47.  If  the  earth  revolve  0°  15'  per  minute,  how  far  per  hour  ? 

48.  Multiply  39A.  3R.  17p.  30yd.  8ft.  lOOin.  by  32. 

49.  If  a  man  be  2d.  5h.  17m.  19sec.  in  walking  1  degree, 
how  long  would  it  take  him  to  walk  round  the  earth,  allowing 
365£  days  to  a  year  ? 

CASE  III. 

When  the  multiplier  is  such  a  number  as  cannot  be  pro- 
duced by  the  product  of  two  or  more  numbers,  we  should  pro- 
ceed as  in  the  following 

EXAMPLE. 

50.  What  is  the  value  of  53  tons  of  iron,  at  18^.  17s.  lid.  a 
ton? 

£.         8.      d.  £.      a.     d. 

18     17  11                                     18  17  11 
5  3 

94      97  =  price  of  5  tons.      56  13    9  =  price  of  3  tons. 

10 

944     15  10  =  price  of  50  tons.       Because   53    is  a   prime 

56     13     9  =  price  of  3  tons,     number,  that  is,  it  cannot  be 

TTwvi n — ™          •        c-  KO  A         produced  by  the  product  of 

9    7  =  price  of  53  tons.  £ny  ,wo  nu*bers  ^  ^ 

fore  find  a  convenient  composite  number  less  than  the  given 
number,  viz.  50,  which  may  be  produced  by  multiplying  5  by 
10.  Having  found  the  price  of  50  tons  by  the  last  Case,  we 
then  find  the  price  of  the  3  remaining  tons  by  Case  I.,  and 
add  it  to  the  former,  making  the  value  of  the  whole  quantity 
100L£.  9s.  7d. 


SECT,  xiv.]  BILLS.  81 

The  pupil  will  hence  perceive  the  propriety  of  the  following 

RULE. 

Take  for  successive  multipliers  two  or  more  numbers,  whose  continued 
product  will  be  nearest  the  proper  multiplier,  and  then  find  the  value  of 
the  remainder  by  Case  I. ,  and  the  sum  of  the  last  two  products  will  be 
the  answer. 

.     51.  What  will  57  gallons  of  wine  cost,  at  8s.  3|d.  per  gallon  ? 

52.  Bought  29  lots  of  wild  land,  each  containing  117A.  3R. 
27p. ;  what  were  the  contents  of  the  whole  ? 

53.  Bought  89  pieces  of  cloth,  each  containing  37yd.  3qr. 
2na.  2in. ;  what  was  the  whole  quantity  ? 

54.  Bought  59  casks  of  wine,  each  containing  47gal.  3qt.  Ipt. ; 
what  was  the  whole  quantity  ? 

55.  If  a  man  travel  17m.  3fur.  13rd.  14ft.  in  one  day,  how 
far  will  he  travel  in  a  year  ? 

56.  If  a  man  drink  3gal.  Iqt.  Ipt.  of  beer  in  a  week,  how 
much  will  he  drink  in  52  weeks  ? 

57.  There  are   17  sticks  of  timber,  each    containing  37ft. 
978in. ;  what  is  the  whole  quantity  ? 

58.  There    are  17  piles  of  wood,  each  containing  7  cords 
98  cubic  feet ;  what  is  the  whole  quantity  ? 

59.  Multiply  2hhd.  19gal.  Oqt.  Ipt.  by  39. 

60.  Multiply  3bu.  Ipk.  4qt.  Ipt.  Igi.  by  53. 

61.  Multiply  16ch.  7bu.  2pk.  Oqt.  Opt.  by  17. 

BILLS. 

1.  London,  July  4,  1835. 

Dow,  Vance,  &  Co.,  of  Boston,  U.  S., 

Bought  of  Samuel  Snow, 
45  yds.  Broadcloth,  at         8s.     4d. 

50    "  "  "       10s.     6d. 

56    "  "  "         3s.     7£d. 

63    "  "  "       12s.  llfd. 

72    "  "  "       19s.   lid. 

81    "  "  "         9s.     3d. 

35    "  "  "       19s.     7£d. 

99    "          "  "       16s.    0£d. 

66    "          "  "        8s.  lid. 

33    "          «  "       16s.  ll£d. 

~376£.  7s.  Ofd. 
Received  payment, 

Samuel  Snow. 


>  COMPOUND  MULTIPLICATION.  [SECT.  xiv. 

2.  Quebec,  Jan.  8,  1835. 
Mr.  John  Vose,                     Bought  of  Vans  &  Conant, 

46  Ivory  Combs,  at         3s.     5£d. 

47  Ibs.  Colored  Thread, "         6s.     9£d. 

51  yds.  Durant,  "         Is.     8d. 

52  Silk  Vests,  "         6s.     7d. 

53  Leghorns,  "        11s.     9£d. 

57  ps.  Nankin,  "         8s.     3±d. 

58  Ibs.  White  Thread,    "         9s.  lljd. 

~128£.  16s. 

Received  payment,      Vans  &  Conant. 

3.  Montreal,  July  4,  1835. 
Mr.  James  Savage,                 Bought  of  Joseph  Dowe, 

83  gals.  Lisbon  Wine,      at  6s.  7d. 

85  "      Port         do.          "  3s.  9£d. 

86  "      Madeira  do.          "  4s.  ll£d. 

87  "     Temperance  do.  "  3s.  6jd. 
89    "      Oil,                       "  5s.  3d. 

91  Leghorns,  "     19s.  10£d. 

92  Ibs.    Green  Tea,          "       3s.     l£d. 

93  pair  Thread  Hose,       "       4s.     4£d. 

94  «      Silk  Gloves,         "       3s.     3£d. 

95  "      Silk  Hose,  "       6s.     6£d. 

97  yds.    Linen,  "       5s.     5]d. 

98  gals.  Winter  Strained  Oil,    7s.     7^d. 

~338<£.  19s. 
Received  payment,  Joseph  Dowe. 

4.  Montreal,  June  17,  1835. 
Mr.  Samuel  Simpson,    Bought  of  Lackington,  Grey,  &  Co. 

19  yds.  Cloth,  at 

23  "  Worsted,  " 

26  "  Baize,  " 

29  u  Camlet,  " 

31  "  Bombazine,  " 

34  "  Linen,  " 

37  "  Cotton,  " 

38  «  Flannel,  " 

39  "  Calico,  " 
41  "  Broadcloth,  " 
43  «  Nankin,  " 

Received  payment,        Lackington,  Grey,  &  Co. 


SECT,  xiv.]  BILLS.  * 

5.  Liverpool,  June  2,  1835. 

John  Jones,  of  Philadelphia,  U.  S., 

Bought  of  Thomas  Hasseltine, 
297  yds.  Black  Broadcloth,     at     17s.     3£d. 
473    "    Blue          do.  "       9s. 

512    "    Red  do.  "     15s.  H 

624    "    Green        do.  "     12s.     8d. 

765    "    White        do.  "     19s.     9£d. 

169    "    Black  Velvet,  <c     13s.     5£d. 

698    "    Green      do.  "     15s.     6fd. 

315    "    Red          do.  "     14s.     3£d. 

713    "    Wliite      do.  "     11s.     7£d. 

519    "    Carpet,  "     13s.     6£d. 

147    "    Black  Kerseymere,  "     16s.     7|d. 
386    "    Blue  do.          "     14s.     3±d. 

137    "    Green          do.          "     19s.     9d. 
999    "    Black  Silk,  "     15s.     8d. 

5012^.  Os. 
Received  payment, 

Thomas  Hasseltine. 


6.  London,  May  11,  1846. 

Messrs.  Kimball,  Jewett,  &  Co.,  of  Boston,  U.  S., 

Bought  of  Benjamin  Fowler, 

£.      s.     d. 
2345  yds.  Red  Broadcloth,  at     l£.  17s.     9£d.  =    4428     12 


7186 
8011 
6789 
3178 


Green  do.  "  3£.  15s.  8£d.  =  27202  0  1 

Black  do.  "  2£.  18s.  10|d.  =  23574  0  8| 

Blue  do.  "  l£.  6s.  9d.    =     9080  5  9 

White  do.  "  2£.  Is.  7|d.  =     6617  10  5<| 


2365  '  Pongee  Silk,  "  l£.     2s.  8£d.  =    2685  5  2£ 

5107  «  Black      do.  '«  l£.     7s.  5^d.  =    7006  3  3| 

4444  bales  Cotton  Cloth,  "  3£.  16s.  8^d.  =  17039  19  3 

7777  "  Irish  Linen,  "  17£.   19s.  9d.    =139888  15  9 

1234  "  Nankin,  "  7£.  15s.  lid.    =    9620  1  2 

4567  "  Flannel,  "  8£.  16s.  7^d.  =  40332  6  4i 

9876  "  Bombazine,  "  3£.     5s.  5|d.  =  32333  12  3 

7658  "  Calico,  "  9£.  17s.  6^d.  =  75638  14  1 

8107  "  Camlet,  "  S£.     6s.  0|d.  =  67313  8  8| 

4725  "  Baize,  '"  3£.     7s.  9fd.  =  16020  14  0| 

5670  "  Durant,  4<  4£.  18s.  5|d.  =  27907  0  7i 


506688  £.  10s.  4|d. 
Received  payment, 

Benjamin  Fowler. 


84  COMPOUND  DIVISION  [SECT.  xv. 

SECTION  XV. 
COMPOUND  DIVISION. 

COMPOUND  DIVISION  is  when  the  dividend  consists  of  several 
denominations. 

EXAMPLES. 

1.  Divide  59&£.  8s.  9d.  equally  among  5  persons. 

£.  fl<  d.  Having  divided  the  pounds  by  5,  we  find 
5)598  8  9  3£.  remaining,  which  are  60s. ;  to  these  we 

TJQ — TO — Q     add  the  8s.  in  the  question,  and  again  divide 

by  5  and  find  3s.  remaining,  which  are  36d. ; 

to  these  we  add  the    9d.  in  the  question,  and  divide  their  sum 

by  5.     The  several  quotients  we  write  under  their  respective 

denominations. 

2.  Divide  168<£.  15s.  Od.  equally  among  36  men. 

I'K.     fit  A  £        When  the  divisor  is  more  than  12,  we 

(**"  usually   perform  the  operation  by  Long 

Division.     In  the  present  example,   we 

24  first  divide  the  pounds  by  36,  and  obtain 

4£.  for  the  quotient  and  24 £.  remaining, 

36)495(13s.  which  we  reduce  to  shillings  and  annex 

36  the  15s.,  and  again  divide  by  36,  and  ob- 

"135  tain  13s.  for  the  quotient.    The  remainder 

108  we  reduce  to  pence,  and  again  divide,  and 

-5=  obtain  9d.  for  the  quotient. 

Zl 

12 

36)324(9d. 
324 

From  the  above  examples  and  illustrations  we  deduce  the 
following 

RULE. 

Divide  tlie  highest  denomination  of  the  dividend  by  the  divisor,  and, 
if  there  be  a  remainder,  reduce  it  to  the  next  lower  denomination,  adding 
to  the  number  thus  found  the  number  in  the  dividend  of  the  same  de- 
nomination. Divide  the  result  thus  obtained  by  the  divisor ;  and,  if 
there  be  a  remainder,  proceed  as  before,  till  all  the  denominations  of  tlie 
dividend  are  taken,  or  till  the  work  is  finished.  The  successive  quo- 
tients will  be  of  the  same  denominations  with  the  successive  numbers 
divided,  or  will  correspond  with  the  several  denominations  of  the  divi- 
dend. 


SECT,  xxi  ]  VULGAR  FRACTIONS.  133 

66.  If  17^  bushels  of  wheat  sow  7T4T  acres,  how  many  bush- 
els will  it  require  to  sow  1  acre  ?  Ans.  2f  bushels. 

67.  John  Jones  sold  his  horse  for  $  176.18.     He  received  in 
part  pay  47f  bushels  of  rye  at  $1.37£  per  bushel ;  and  for  the 
remainder  he  received  wheat  at  $  2^  per  bushel.    Required  the 
quantity  of  wheat.  Ans.  45T6^(y3-(J  bushels. 

68.  Sold  apples  at  f  of  a  dollar  a  bushel.     What  did  I  re- 
ceive for  SI  bushels  ?  Ans.  $  1.17f . 

69.  Sold  18f  bushels  of  rye  for  $21.47.  What  was  received 
for  1  bushel  ?    what  for  7£  bushels?  Ans.  $  9.19^. 

70.  Gave  17£.  8s.  lid.  for  9T3T  tons  of  coal.     What"  cost 
1  ton?    what  cost  19f  tons  ?  Ans.  36<£.  16s.  5|f  Jd. 

71.  How  many  garments  can  be  made  of  756T4T  yards  of 
cloth,  if  each  garment  requires  7-f  |  yards  ? 

Ans.  100  garments. 

72.  If  18f  cords  of  wood  cost  $  90.00,  what  cost  1  cord  ? 
what  cpst  171f  cords  ?  Ans.  $  830.50££. 

73.  If  a  man  travel  147  r3T  miles  in  36|  hours,  how  far  will 
he  travel  in  1  hour  ?    how  far  in  97£  hours  ?    Ans.  392f  |||m. 

74.  If  a  man  travel  500  miles  in  97f  hours,  how  far  will  he 
travel  in  32^  hours  ?  Ans.  166§  miles. 

75.  If  a  horse  eat  19f  bushels  of  oats  in  87f  days,  how 
many  will  7  horses  eat  in  60  days  ?  Ans.  93^  bushels. 


MISCELLANEOUS   QUESTIONS. 

1.  How  far  will  a  man  walk  in  17T3T  hours,  provided  he 
goes  at  the  rate  of  4£  miles  an  hour  ? 

Ans.  82m.  4fur.  8rd.  1ft.  4in. 

2.  How  much  land  is  there  in  a  field  which  is  29T77j-  rods 
square  ?  Ans.  5A.  1R.  32p.  141ft.  109-f£f  in. 

3.  How  much  wood  in  a  pile  which  is  17f  feet  long,  7T\. 
feet  high,  and  4f  feet  wide  ?  Ans.  4cd.  66||^ft. 

4.  What  is  the  value  of  19 £  barrels  of  flour,  at  $  6f  a  bar- 
rel ?  ,  Ans.  $134.15f. 

5.  What  is  the  value  of  376|£  acres  of  land,  at  $75f  per 
acre  ?  Ans.  $  28387.06^-. 

6.  What  cost  17^  quintals  of  fish,  at  $  4.75  per  quintal  ? 

Ans.  $  81.55-&V. 

7.  What  cost  1670^  pounds  of  coffee,  at  12f  cents"  per 
pound?  Ans.  $212.99|f. 


134  VULGAR  FRACTIONS.  [SECT.  xxi. 

8    What  cost  28T4T  tons  of  Lackawana  coal,  at  $1  If  a  ton  ? 

Ans.  $  333.27T3T. 

9.  Bought  37££  hogsheads  of  molasses,  at  $  17.62£  a  hhd. ; 
what  was  the  whole  cost  ?  Ans.  $  655.20/T. 

10.  What  cost  £  of  a  cord  of  wood,  at  $  5.75  a  cord  ? 

Ans.  $5.03*, 

11.  What  are  the  contents  of  a  field  which  is  1394  rods 
long,  and  38±  rods  wide  ?  Ans.  33A.  3R.  15£f  p. 

12.  Bought  15  loads  of  wood,  each  containing  llf  feet,  cord 
measure.     I  divide  it  equally  between  9  persons ;  what  does 
each  receive  ?  Ans.  19£ft. 

13.  If  the  transportation  of  18f  tons  of  iron  cost  $  48.15|-, 
what  is  it  per  ton  ?  Ans.  $  2.62^. 

14.  If  a  hhd.  of  wine  cost  $  98|,  what  is  the  price  of  one 
gallon  ?  Ans.  $  1.56|Jf. 

15.  If  5  bushels  of  wheat  cost  $  8|,  what  will  a  bushel  be 
worth?  Ans.  $  1.64£. 

16.  What  will  11  hogsheads  and  17 £  gallons  of  wine  cost, 
at  19f  cents  a  gallon  ?  Ans.  8  140.32$. 

17.  How  many  bottles,  each  containing  1 J  pints,  are  suffi- 
cient for  bottling  a  hhd.  of  cider  ?  Ans.  288. 

18.  I  have  a  shed  which  is  18^  feet  long,  10TV  feet  wide, 
and  7|4  feet  high  ;  how  many  cords  of  wood  will  it  contain  ? 

Ans.  lied.  124#?ftft. 

19.  What  will  6%  pounds  of  tea  cost,  at  65£  cents  per  Ib.  ? 

Ans.  $  4.52?V- 

20.  How  many  cubic   feet  does  a  box  contain,  that  is  8£ 
feet  long,  5/y  feet  wide,  and  3  feet  high  ?         Ans.  146T9^ft. 

21.  How  many  feet  of  boards  will  it  take  to  cover  a  side 
of  a  house  which  is  46TV  feet  long  and  17£  feet  high  ? 

Ans.  812^ft. 

22.  Required  the  number  of  square  feet  on  the  surface  of  7 
boxes,  each  of  which  is  5£  feet  long,  2^  feet  high,  and  3^ 
feet  wide  ;  required  also  the  number  of  cubic  feet  they  would 
occupy.  Ans.  527£f  ft. ;  286f  f  f  cubic  feet. 

23.  A  certain  room  is  12  feet  long,  ll£  feet  wide,  and  7£ 
feet  high  ;  how  much  will  it  cost  to  plaster  it,  at  2f  cents  per 
square  foot  ?  Ans.  $  13.48|-. 

24.  A  man  has  a  garden  that  is  14£  rods  long,  and  10£  rods 
wide  ;  he  wishes  to  have  a  ditch  dug  around  it,  that  shall  be 
3  feet  wide  and  4£  feet  deep ;  what  will  be  the  expense,  if  he 
gives  2  cents  per  cubic  foot  ?  Ans.  $  223.76£. 

25.  How  many  bushels  of  grain  will  a  box  contain  which  is 


SECT,  xxii.]  DECIMAL  FRACTIONS.  135 

14T72-  feet  long,  5T£  feet  deep,  and  4£  feet  wide,  there  being 
2150f  cubic  inches  in  a  bushel  ?  Ans.  294ff ffbu. 

26.  Which  will  contain  the  most,  and  by  how  much,  a  box 
that  is  10  feet  long,  8  feet  wide,  and  6  feet  deep,  or  a  cubical 
one,  whose  each  side  measures  8  feet  ? 

Ans.  The  last  contains  32  cubic  feet  the  most. 

27.  Divide  $  1 1 12f  equally  among  129  men.    Ans.  $  8  -£. 

28.  Bought  68  barrels  of  flour,  at  $  7T£  per  barrel ;  what  was 
the  amount  of  the  whole  ?  Ans.  $  538£. 

29.  What  cost  8f  acres  of  land,  at  $  42f  per  acre  ? 

Ans.  f  369.20. 

30.  How  shall  four  3's  be  arranged,  that  their  value  shall  be 
nothing  ? 

31.  I  have  a  room  20  feet  long,  15  feet  wide,  and  8£  feet 
high.     This  room  contains  4  windows,  each  of  which  is  5^  feet 
in  height  and  3£  feet  in  width.     There  are  two  doors  7  feet 
high  and  3  feet  wide.     The  mop-boards  are  f  of  a  foot  wide. 
A  mason,  has  agreed  to  plaster  this  room  at  6£  cents  per  square 
yard  ;  a  painter  is  to  lay  on  the  paper  at  9  cents  per  square 
yard ;  the  paper  which  I  wish  to  have  laid  on  is  2f  feet  wide, 
for  which  I  pay  5  cents  per  yard.     What  is  the  amount  of  my 
bill  for  plastering,  for  papering,  and  for  paper  ?  .' 

Ans.  For  plastering,  $  5.1  Iff ;  for  papering,  8  4.37  ;  for  pa- 
per, $  2.80  A- 


SECTION  XXII. 
DECIMAL   FRACTIONS. 

A  DECIMAL  FRACTION  is  a  fraction  whose  denominator  is  a 
unit  with  as  many  ciphers  annexed  as  there  are  places  in  the 
numerator.  Thus,  y5^,  yVfr,  r^V  &c- 

The  denominator  being  in  all  cases  formed  in  this  obvious 
manner,  there  is  no  necessity  of  expressing  it,  and  therefore  it 
is  not  written  ;  but  a  point  is  placed  before  the  first  figure  of 
the  numerator,  to  indicate  that  the  figure  or  figures  on  the 
right  of  the  point  denote  a  decimal,  and  not  a  whole  number. 
Thus,  for  T5^  we  write  .5,  for  yVfr,  .15,  &c.  This  point  is 
called  the  separatrix. 

Ciphers  annexed  to  the  right  hand  of  decimals  do  not  in- 
crease their  value  ;  for  .4,  and  .40,  and  .400  are  decimals  having 


X36  DECIMAL  FRACTIONS.  [SECT.  xxn. 

the  same  value,  each  being  equal  to  -fV  or  f  ;  but  when  ciphers 
are  placed  on  the  left  hand  of  a  decimal,  they  decrease  the 
value  in  a  tenfold  proportion.  Thus,  .4  is  y4^,  or  four  tenths  ; 
but  .04  is  y4^,  or  four  hundredths  ;  and  .004  is  y^4^,  or  four 
thousandths.  The  figure  next  the  separatrix  is  reckoned  so 
many  tenths  ;  the  next  at  the  right,  so  many  hundredths  ;  the 
third  is  so  many  thousandths  ;  and  so  on,  as  may  be  seen  by 
the  following 


TABLE. 


I 


6 


I    I 

1 1 1 1  i 

II  I  P  I  1 

0         P         <-;  3       r^3 

H    ffi    h«    H    K    5S 
7654321.    234567 

From  this  table  it  is  evident,  that  in  decimals,  as  well  as  in 
whole  numbers,  each  figure  takes  us  value  by  its  distance  from 
the  place  of  units. 

NOTE.  —  If  there  be  one  figure  in  the  decimal,  it  is  so  many  tenths; 
if  there  be  two  figures,  they  express  so  many  hundredths ;  if  there  be 
three  figures,  they  are  so  many  thousandths,  &c. 

NUMERATION  OF  DECIMAL  FRACTIONS. 

Let  the  pupil  write  the  following  numbers. 

1.  Three  hundred  twenty-five,  and  seven  tenths. 

2.  Four  hundred  sixty-five,  and  fourteen  hundredths. 

3.  Ninety-three,  and  seven  hundredths. 

4.  Twenty-four,  and  nine  millionths. 

5.  Two  hundred  twenty-one,  and  nine  hundred  thousandths. 

6.  Forty-nine  thousand,  and  forty-nine  thousandths. 

7.  Seventy-nine  million  two  thousand,  and  one  hundred  five 
thousandths. 

8.  Sixty-nine  thousand  fifteen,  and  fifteen  hundred  thou- 
sandths. 

9.  Eighty  thousand,  and  eighty-three  ten  thousandths. 


SECT,  xxni.]  DECIMAL  FRACTIONS.  137 

10.  Nine  billion  nineteen  thousand  nineteen,  and  nineteen 
hundredths. 

11.  Twenty-seven,   and   nine   hundred,  twenty -seven   thou- 
sandths. 

12.  Forty-nine  trillion,  and  one  trillionth. 

13.  Twenty-one,  and  one  ten  thousandth. 

14.  Eighty-seven  thousand,  and  eighty-seven  millionths. 

15.  Ninety-nine    thousand  ninety-nine,  and   nine  thousand 
nine  billionths. 

16.  Seventeen,  and  one  hundred  seventeen  ten  thousandths. 

17.  Thirty-three,  and  thirty-three  hundredths. 

18.  Forty-seven  thousand,  and  twenty-nine  ten  millionths.    • 

19.  Fifteen,  and  four  thousand  seven  hundred  thousandths. 

20.  Eleven  thousand,  and  eleven  hundredths. 

21.  Seventeen,  and  eighty-one  quatrillionths. 

22.  Nine,  and  fifty-seven  trillionths. 

23.  Sixty-nine  thousand,  and  three  hundred  forty-nine  thou- 
sandths. 

Let  the  following  numbers  be  written  in  words. 

27.86  86.0007  1.000007  16.300000007 

48.07  5.6001  5.101016  1.315 

15.716  34.1063  6.716678  0.0000001 

161.3  15.0016  1.631  10.10101 

87.006  16.1004  3.760701  1.000327 


SECTION  XXIII. 
ADDITION  OF  DECIMALS. 

1.  Add  23.61  and  161.5  and  2.6789  and  61.111  and  27.0076 
and  116.71  and  6151.7671  together. 

OPERATION. 

23.61  In  this  question,  it  will  be  perceived 

161.5  that    tenths   are   written    under  tenths, 

2.6789  hundredths  under  hundredths,  &c. ;  and 

61.111  that   the  operation   of  addition   is   per- 

27.0076  formed   as  in  addition  of  whole  num- 

116.71  bers. 
6151.7671 

6544.3846 

12* 


138  DECIMAL  FRACTIONS.  [SECT.  xxir. 

RULE.  —  Write  the  numbers  under  each  other  according  to  their 
value,  add  as  in  whole  numbers,  and  point  off  from  the  right  hand  as 
many  places  for  decimals  as  there  are  in  that  number  which  contains 
the  greatest  number  of  decimals. 

2.  Add  together  the  following  numbers  :  81.61356,  6716.31, 
413.1678956,  35.14671,  3.1671,  314.6.   Ans.  7564.0052656. 

3.  What  is  the  sum  of  the  following  numbers  :    1121.6116, 
61.87,  46.67,  165.13,  676.167895  ?  Ans.  2071.449495. 

4.  Add  7.61,  637.1,  6516.14,  67.1234,  6.1234,  together. 

Ans.  7234.0968. 

5.  Add  21.611,  6888.32,  3.6167,  together. 

Ans.  6913.5477. 

6.  Add  seventy-three  and  twenty-nine  hundredths,  eighty- 
seven  and  forty-seven  thousandths,  three  thousand  and  five  and 
one  hundred  six  ten  thousandths,  twenty-eight  and  three  hun- 
dredths, twenty-nine  thousand  and  five  thousandths,  together. 

Ans.  32193.3826. 

7.  Add  two  hundred  nine  thousand  and  forty-six  millionths, 
ninety-eight  thousand  two  hundred  seven  and  fifteen  ten  thou- 
sandths, fifteen  and  eight  hundredths,  and  forty -nine  ten  thou- 
sandths, together.  Ans.  307222.086446. 

8.  What  is  the  sum  of  twenty-three  million  ten,  one  thou- 
sand and  five  hundred  thousandths,  twenty-seven  and  nineteen 
millionths,  seven  and  five  tenths  ?      Ans.  23001044.500069. 

9.  Add   the   following   numbers  :    fifty-nine   and   fifty-nine 
thousandths,  twenty-five   thousand  and   twenty- five  ten  thou- 
sandths, five  and  five  millionths,  two  hundred  five  and  five 
hundredths.  Ans.  25269.1 1 1505. 

10.  What  is  the  sum  of  the  following  numbers :  twenty-five 
and  seven  millionths,  one  hundred  forty-five  and  six  hundred 
forty-three  thousandths,  one  hundred  seventy-five  and  eighty- 
nine  hundredths,  seventeen  and  three  hundred  forty-eight  hun- 
dred thousandths  ?  Ans.  363.536487. 


SECTION  XXIV. 
SUBTRACTION  OF  DECIMALS. 

RULE.  —  Let  the  numbers  be  so  written,  that  the  separatrix  of  the 
subtrahend  be  directly  under  that  of  the  minuend,  subtract  as  in  whole 


SECT.  XXV.] 


DECIMAL  FRACTIONS. 


139 


numbers,  and  point  off  a&  many  places  for  decimals  as  there  are  in  that 
number  which  contains  the  greatest  number  of  decimals. 


1. 

61.9634 
9.182 

52.7814 


OPERATIONS. 
2. 

39.3 
1.6789 


37.6211 


3. 

5. 

1.678 

3.322 


4.10001 


5.  From  41.7  take  21.9767.  Ans.  19.7233. 

6.  From  29.167  take  19.66711.  Ans.  9.49989. 

7.  From  91.61  take  2.6671.  Ans.  88.9429. 

8.  From  96.71  take  96.709.  Ans.  .001. 

9.  Take  twenty-seven  and  twenty-eight  thousandths  from 
ninety-seven  and  seven  tenths.  Ans.  70.672. 

10.  Take  one  hundred  fifteen  and  seven  hundredths  from 
three  hundred  fifteen  and  twenty-seven  ten  thousandths. 

Ans.  199.9327. 

11.  From  twenty-nine  million  four  thousand  and  five  take 
twenty-nine  thousand  and  three  hundred  forty-nine  thousand 
two  hundred,  and  twenty-four  hundred  thousandths. 

Ans.  28625804.99976. 

12.  From  one  million  take  one  millionth. 

Ans.  999999.999999. 


SECTION  XXV. 
MULTIPLICATION  OF  DECIMALS. 


EXAMPLES.         OPERATIONS.     VULG.  FRAC.     DECIMALS. 


-fr 


1.  Multiply   A   b7   A- 

2.  Multiply 

3.  Multiply 

4.  Multiply 

5.  Multiply 

6.  Multiply 

7.  Multiply  TT/ 

8.  Multiply  ^A  by  T 

9.  Multiply  TVirV  by  32. 
10.  Multiply  Tfo  by  46. 


b7 
by 
by 
by 
by 
by 
by 


AxA   = 

AXA  = 


=  w 


=1     .56 

=  .49 
=  .036 
=  .0008 
=  .00063 
=  .335284 
=21.472 
=  3.22 


140                               DECIMAL  FRACTIONS.  [SECT.  ixv. 

11.  Multiply  76.81  by  3.2.  12.  Multiply  .1234  by  .0046. 

OPERATION.  OPERATION. 

76.81  .1234 
3.2  .0046 

15362  7404 

23043  4936 


245.792  .00056764 

RULE.  —  Multiply  as  in  whole  numbers,  and  point  off  as  many  fig- 
ures for  decimals  in  the  product  as  there  are  decimals  in  the  multipli- 
cand and  multiplier ;  but,  if  there  should  not  be  so  many  figures  in  the 
product  as  there  are  decimals  in  the  multiplicand  and  multiplier,  supply 
the  defect  by  prefixing  ciphers.  See  Example  12th. 

13.  Multiply  61.76  by  .0071.  Ans.  .438496. 

14.  Multiply  .0716  by  1.326.  Ans.  .0949416. 

15.  Multiply  .61001  by  .061.  Ans.  .03721061. 

16.  Multiply  71.61  by  365.  Ans.  26137.65. 

17.  Multiply  .1234  by  1234.  Ans.  152.2756. 

18.  Multiply  6.71 1  by  6543.  Ans.  439 10.073. 

19.  Multiply  .0009  by  .0009.  Ans.  .00000081. 

20.  Multiply  forty-nine  thousand  by  forty-nine  thousandths. 

Ans.  2401. 

21.  What  is  the  product  of  one  thousand  and  twenty-five, 
multiplied  by  three  hundred  and  twenty-seven  ten  thousandths  ? 

Ans.  33.5175. 

22.  What  is  the  product  of  seventy-eight  million  two  hun- 
dred  five    thousand    and   two,  multiplied  by  fifty-three  hun- 
dredths  ?  Ans.  41448651.06. 

23.  Multiply  one  hundred  and  fifty-three  thousandths  by  one 
hundred  twenty-nine  millionths.  Ans.  .000019737. 

24.  What  is  the  product  of  fifteen  thousand,  multiplied  by 
fifteen  thousandths  ?  Ans.  225. 

25.  What  will  26.7  yards  of  cloth  cost,  at  $  5.75  a  yard  ? 

Ans.  $153.52,5. 

26.  What  will  14.75  bushels  of  wheat  cost,  at  81.25  a  bushel  ? 

Ans.  $18.43,75. 

27.  What  will  375.6  pounds  of  sugar  cost,  at  $0.125  per 
pound  ?  Ans.  $  46.95. 

28.  What  will   26.58   cords   of  wood   cost,   at   $5.625   a 
cord?  Ans.  $  149.5 1,2£. 

29.  What  will  28.75  tons  of  potash  cost,  at  $125.78~per 
ton?  Ans.  $3616.17,5. 


SECT,  xxvi.]  DECIMAL  FRACTIONS.  141 

30.  What  will  369  gallons  of  molasses  cost,  at  $0.375  a 
gallon?  Ans.  $138.37,5. 

31.  What  will  97.48cwt.  of  hay  cost,  at  $1.125  per  cwt.  ? 

Ans.  $109.66,5. 

32.  What  will  63.5  bushels  of  corn  be  worth,  at  $  0.78  per 
bushel?  Ans.  $49.53. 


SECTION  XXVI. 
DIVISION  OF  DECIMALS. 

EXAMPLES.  OPERATIONS.          VULG.  FRAC.     DECIMALS. 

1.  Divide  A  by    T2^.  T8<yX-V°-     =     f§  =  «  =  4. 

2.  Divide  -fa  by   TV  T3zyX-\p-     =     f %  =  ffr  =  .5 

3.  Divide  f^  by    TV  T3<&X-V°-     =  f*ft  =  A  =  -9 

4.  Divide  ±f  by     4.  ffxi       =     ff  =  if  ==  1.2 

5.  Divide  -^  by  TW  fAx^0-  =ff$$  =  f §  =  8. 

6.  Divide  yfo.  by  y^.  T$*XJ^*==^W  =  •*£  =  10. 

7.  Divide  1.728  by  1.2.  8.  Divide  |&§f  by  if. 

OPERATION  BY  DECIMALS.       OPERATION  BY  VULGAR  FRACTIONS. 

1.2)  1.728(1.44  Ans. 
12 
52 

48 

48 
48 

RULE.  —  Divide  as  in  whole  numbers,  and  point  off  as  many  deci- 
mals in  the  quotient  as  the  number  of  decimals  in  the  dividend  exceed 
those  of  the  divisor ;  but  if  the  number  of  those  in  the  divisor  exceed  that 
of  the  dividend,  supply  the  defect  by  annexing  ciphers  to  the  dividend. 
And  if  the  number  of  decimals  in  the  quotient  and  divisor  together  are 
not  equal  to  the  number  in  the  dividend,  supply  the  defect  by  prefixing 
ciphers  to  the  quotient. 

9.  Divide  780.516  by  2.  43.  Ans.  321.2. 

10.  Divide  7.25406  by  9.57.  Ans.  .758. 

11.  Divide  .21318  by  .38.  Ans.  .561. 

12.  Divide  7.2091365  by  .5201.  Ans.  13.861+. 

13.  Divide  56.8554756  bv  .0759.  Ans.  749.084. 


142  DECIMAL  FRACTIONS.  [SECT.  xxvn. 

14.  Divide  30614.4  by  .9567.  Ans.  32000. 

15.  Divide  .306144  by  9567.  Ans.  .000032. 

16.  Divide  four  thousand  three  hundred  twenty-two  and  four 
thousand  five  hundred  seventy-three  ten  thousandths  by  eight 
thousand  and  nine  thousandths.  Ans.  .5103+. 

17.  Divide  thirty-six  and  six  thousand  nine  hundred  forty- 
seven  ten  thousandths  by  five  hundred  and  eighty-nine. 

Ans.  .0623. 

18.  Divide  three  hundred  twenty-three  thousand  seven  hun- 
dred sixty- five  by  five  mill  ion  ths.    '  Ans.  64753000000. 

19.  Divide  119109094.835  by  38123.45.         Ans.  3124.3. 

20.  Divide  1191090.94835  by  3812345.          Ans. 

21.  Divide  11910909483.5  by  38.12345.        Ans. 

22.  Divide  11.9109094835  by  381234.5.         Ans. 

23.  Divide  1191.09094835  by  3.812345.         Ans. 

24.  Divide  11910909483.5  by  .3812345.         Ans. 

25.  Divide  1.19109094835  by  3.812345.         Ans. 

26.  Divide  .119109094835  by  .3812345.        Ans. 


SECTION  XXVII. 
REDUCTION  OF  DECIMALS. 

CASE  I. 

To  reduce  a  vulgar  fraction  to  its  decimal. 
1.  Reduce  f  to  its  decimal. 

OPERATION.  That  the  decimal  .375  is  equal  to  f  may  be 

8)3.000  shown  by  writing  it  in  a  vulgar  fraction  and  re- 

ducing  h  5  thus» 


RULE.  —  Divide  the  numerator  by  the  denominator,  annexing  one  or 
more  ciphers  to  the  numerator,  and  the  quotient  will  be  the  decimal  re- 
quired. 

NOTE.  —  It  is  not  usually  necessary  that  decimals  should  be  carried  to 
more  than  six  places. 

2.  Reduce  £  to  a  decimal.  Ans.  .625. 

3.  Reduce  %  to  a  decimal.  Ans.  .5. 

4.  Reduce  f ,  £ ,  |- ,  J-£,  T3^,  ^  ana  £  to  decimals. 

Ans.  .666+,  .75,  .833+,  .91666+,  .1875,  .04,  .125. 


SECT,  xxvii.]  DECIMAL  FRACTIONS.  143 


4 
12 
20 


5.  Reduce  TV,  fa  fa  f  §T,  rf  T,  and  y^1^  to  decimals. 
Ans.  .05882+,  .07407-)-,  .1351+,  .00696+,  .07207+, 
.0008103+. 

CASE  II. 

To  reduce  denominate  numbers  to  decimals. 
1.  Reduce  15s.  9f  d.  to  the  decimal  of  a  £.     Ans.  .790625. 
OPERATION.  The  3  farthings  are  f  of  a  penny,  and 

3.00  these  reduced  to  a  decimal  are  .75  of  a 

9.75000  penny,  which  we  annex  to  the  pence,  and 

15.81250  proceed  hi  the  same  manner  with  the  other 


.790625  Ans.     terms- 


RULE.  —  Write  the  given  numbers  perpendicularly  under  each  other 
for  dividends,  proceeding  orderly  from  the  least  to  tJie  greatest ;  opposite 
to  each  dividend,  on  the  left  hand,  place  such  a  number  for  a  divisor,  as 
will  bring  it  to  the  next  superior  denomination,  and  draw  a  line  between 
them.  Begin  at  the  highest,  and  write  the  quotient  of  each  division,  as 
decimal  parts,  on  the  right  of  the  dividend  next  below  it,  and  so  on,  till 
they  are  all  divided  ;  and  the  last  quotient  will  be  the  decimal  required. 

2.  Reduce  9s.  to  the  fraction  of  a  pound.  Ans.  .45. 

3.  Reduce  15cwt.  3qr.  141b.  to  the  decimal  of  a  ton. 

Ans.  .79375. 

4.  Reduce  2qr.  2 lib.  8oz.  12dr.  to  the  decimal  of  a  cwt. 

Ans.  .6923828125. 

5.  Reduce  Iqr.  3na.  to  the  decimal  of  a  yard. 

Ans.  .4375. 

6.  Reduce  5fur.  35rd.  2yd.  2ft.  9in.  to  the  decimal  of  a  mile. 

Ans.  .73603219+. 

7.  Reduce  3gal.  2qt.  Ipt.  of  wine  to  the  decimal  of  a  hogs- 
head. Ans.  .0575396+. 

8.  Reduce  Ipt.  to  the  decimal  of  a  bushel.    Ans.  .015625. 

9.  Reduce  2R.  16p.  to  the  decimal  of  an  acre.        Ans.  .6. 

CASE  III. 

To  find  the  decimal  of  any  number  of  shillings,  pence,  and 
farthings,  by  inspection. 

RULE.  —  Write  half  the  greatest  even  number  of  shillings  for  the  first 
decimal  figure,  and  if  the  number  of  shillings  be  odd,  annex  to  the  deci- 
mal the  figure  5.  Then  write  underneath  the  number  of  farthings  con- 
tained in  the  given  pence  and  farthings,  setting  the  left-hand  figure  in 
the  second  place,  if  there  be  more  than  one  figure,  and  the  single  figure 


144  DECIMAL  FRACTIONS.  [SECT.  xxvn. 

in  the  third  place,  if  there  be  but  one,  and  increasing  the  number  by  I 
wften  it  exceeds  12,  and  by  2  token  it  exceeds  36.  T/ie  sum  of  the  whole 
will  be  the  decimal  required. 

EXAMPLES. 

1.  Find  the  decimal  of  15s.  9£d.  by  inspection. 

.7      =  £  of  14s. 
.05    =  for  odd  shilling. 
39  =  farthings  in  9fd. 
2  =  for  excess  of  36. 

~779l 

2.  Find  the  value  of  13s.  6|d.  by  inspection.      Ans.  .678. 

3.  Find  the  value  of  19s.  8£d.  by  inspection.      Ans.  .984. 

4.  Value  the  following  sums  by  inspection,  and  find  their  to- 
tal:    19s.  llfd.,  16s.  9£d.,  Is.  lid.,  3s.  OJd.,  17s.  5£d.,  13s. 
4£d.,  18s.  8£d.,  19s.  ll£d.,  13s.  3^d.,  16s.  0£d.,  17s.  7£d. 

Ans.  7.r " 


NOTE.  —  As  shillings  are  so  many  twentieths  of  a  pound,  it  is  evident, 
that  by  taking  one  half  of  their  number,  we  obtain  their  value  in  tenths 
or  decimals  of  a  pound.  Thus,  16s.  =  &£. 

In  like  manner,  any  number  of  farthings  are  so  many  nine  hundred  six- 
tieths of  a  pound.  So  that,  in  order  to  obtain  their  value  in  the  denomi- 
nation of  pounds,  we  write  the  number  of  farthings  for  the  numerator  and 
960  for  a  denominator,  as  17  farthings  =  $B£.  But,  in  order  to  treat  this 
fraction  decimally,  we  must  raise  the  denominator  to  1000,  which  in  the 
fraction  j^£>.  is  done  by  adding  40  to  the  denominator  and  1  to  the  nume- 


rator, and  in  the  fraction  gf0  by  adding  2  to  the  numerator  and  40  to  the 
denominator.    Q.  £.  D. 

CASE  IV. 

To  find  the  value  of  a  decimal  in  integral  or  whole  numbers. 
1.  What  is  the  value  of  .790625£.  ? 

OPERATION. 

.790625  Now  it  is  evident,  that  .790625£.  expressed 

20  in  terms  of  a  shilling  must  be  the  product  of 

15  812500  .790625  £.  multiplied  by  20,  and  that  to  con- 

J2  tinue  the  reduction  to  the  lowest  terms  we  must 

h  lyRnnnr  multiply  by  the  same  number  as  in  common  re- 

duction. 
4 

aoooooo 

RULE.  —  Multiply  tlie  given  decimal  by  the  number  which  will  bring 
it  to  the  next  lower  denomination,  and  cut  off  for  a  remainder  as  many 
places  on  the  right  as  there  are  places  in  the  given  decimal. 


SECT,  xxvni.]  DECIMAL  FRACTIONS.  145 

Multiply  this  remainder  by  the  number  which  will  bring  it  to  the  next 
lower  denomination,  cutting  off  for  a  remainder  as  before,  and  thus  pro- 
ceed till  the  reduction  is  carried  to  the  denomination  required.  The  sev- 
eral integral  numbers,  standing  at  the  left  hand,  will  be  the  answer 
sought  in  the  different  lower  denominations. 

2.  What  is  the  value  of  .625  of  a  shilling  ?         Ans.  7£d. 

3.  What  is  the  value  of  .6725  of  a  cwt.  ? 

Ans.  2qr.  191b.  5^°z. 

4.  What  is  the  value  of  .9375  of  a  yard  ?      Ans.  3qr.  3na. 

5.  What  is  the  value  of  .7895  of  a  mile  ? 

Ans.  6fur.  12rd.  10ft.  6£f  in. 

6.  What  is  the  value  of  .9378  of  an  acre  ? 

Ans.  3R.  30p.  13ft.  9T9^m. 

7.  Reduce  .5615  of  a  hogshead  of  wine  to  its  value  in  gal- 
lons, &c.  Ans.  35gal.  Iqt.  Opt.  3|ff  gi. 

8.  Reduce  .367  of  a  year  to  its  value  in  days,  &c. 

Ans.  134da.  Ih.  7m.  19£sec. 

9.  What  is  the  value  of  .6923828125  of  a  cwt.  ? 

Ans.  2qr.  2  lib.  8oz.  12dr. 

10.  What  is  the  value  of  .015625  of  a  bushel  ? 

Ans.  1  pint. 

11.  What  is  the  value  of  .55  of  an  ell  English  ? 

Ans.  2qr.  3na. 

12.  What  is  the  value  of  .6  of  an  acre  ?        Ans.  2R.  16p. 


SECTION  XXVIII. 
MISCELLANEOUS  EXAMPLES. 

1.  What  is  the  value  of  7cwt.  2qr.  181b.  of  sugar,  at  $  11.75 
per  cwt.  ?  Ans.  $  90.01, 3££. 

2.  What  cost  19cwt.  3qr.  141b.  of  iron,  at  $  9.25  per  cwt.  ? 

Ans.  $  183.84,3}. 

3.  What  cost  39A.  2R.  15p.  of  land,  at  $  87.37,5  per  acre  ? 

Ans.  $  3459.50,3f f . 

4.  What  would  be  the  expense  of  making  a  turnpike  87m. 
3fur.  15rd.,  at  $  578.75  per  mile  ?  Ans.  $  50595.41¥V- 

5.  What  is  the  cost  of  a  board   18ft.  9in.  long,  and  2ft.  3^-in. 
wide,  at  $  .05,3  per  foot  ?  Ans.  $  2.27,7^. 

6.  Goliath  of  Gath  was  6^  cubits  high  ;  what  was  his  height 
in  feet,  the  cubit  being  1ft.  7.168in.  ?  -       Ans.  10ft.  4.592in. 

13 


146  DECIMAL   FRACTIONS.  [SECT, 

7.  If  a  man  travel  4.316  miles  in  an  hour,  how  long  would 
he  be  in  travelling  from  Bradford  to  Boston,  the  distance  being 
29£  miles  ?  Ans.  6h.  50m.  6sec.+ 

8.  What  is  the  cost  of  5yd.  Iqr.  2na.  of  broadcloth,  at  $  5.62£ 
per  yard  ?  Ans.  $  30.23,4f . 

9.  Bought  17  bags  of  hops,  each  weighing  4cwt.  3qr.  71b.,  at 
$  5.87£  per  cwt. ;  what  was  the  cost  ?    ~  Ans.  $  480.64,8-^. 

10.  Purchased   a   farm,   containing    176A.    3R.   25rd.,   at 
$  75.37£  per  acre  ;  what  did  it  cost  ?     Ans.  $  13334.3Q,8]f . 

11.  What  cost  17625  feet  of  boards,  at  $12.75  per  thou- 
sand? Ans.  $  224.7  J,8£. 

12.  How  many  square  feet  in  a  floor  19ft.  3in.  long,  and 
15ft.  9in.  wide  ?  Ans.  303ft.  27in. 

13.  How  many  square  yards  of  paper  will  it  take  to  cover  a 
room  14ft.  6in.  long,  12ft.  6in.  wide,  and  8ft.  9in.  high  ? 

Ans.  52£yd. 

14.  How  many  solid  feet  in  a  pile  of  wood  10ft.  7in.  long, 
4ft.  wide,  and  5ft.  lOin.  high  ?  Ans.  246T£ft. 

15.  How  many  garments,  each  containing  4yd.  2qr.  3na., 
can  be  made  from  1 12yd.  2qr.  of  cloth  ?  Ans.  24. 

16.  Bought  Igal.  2qt.  Ipt.  of  wine  for  $  1.82  ;  what  would 
be  the  price  of  a  hogshead  ?  Ans.  $70.56. 

17.  Bought  125£yd.  of  lace  for  $  15.06 ;  what  was  the  price 
of  1  yard?  Ans.  $0.12. 

18.  What  cost  17cwt.  3qr.  of  wool,  at  $  35.75  per  cwt.  ? 

Ans.  $  634.56,2£. 

19.  What  cost  7hhd.  47gal.  of  wine,  at  $  87.25  per  hogs- 
head ?  Ans.  $  675.84^. 

20.  How  many  solid  feet  in  a  stick  of  timber  34ft.  9in.  long, 
1ft.  Sin.  wide,  and  1ft.  6in.  deep  ?  Ans.  65.15625ft. 

21.  How  many  cwt.  of  coffee  in  17f-  bags,  each  bag  contain- 
ing 2cwt.  Iqr.  71b.  ?  Ans.  41cwt.  Oqr.  5£lb. 

22.  If  18yd.  Iqr.  of  cloth  cost  $36.50,  what  is  the  price  of 
1  yard  ?  Ans.  $2.00. 

23.  If  $  477.72  be  equally  divided  among  9  men,  what  will 
be  each  man's  share  ?  Ans.  $  53.08. 

24.  A  man  bought  a  barrel  of  flour  for  $  5.375,  7gal.  of  mo- 
lasses for  $1 .78,  9gal.  of  vinegar  for  $1.1875,  Igal.  of  wine  for 
$1.125,  141b.  of  sugar  for  $1.275,  and  51b.  of  tea,  for  $2.625; 
what  did  the  whole  amount  to  ?  Ans.  $  13.36,7£. 

25.  A  man  purchased  3  loads  of  hay  ;  the  first  contained  2f 
tons,  the  second  3|  tons,  and  the  third   ly1^  tons  ;  what  was  the 
value  of  the  whole,  at  $  17.625  a  ton  ?     Ans.  $  128.88,2T£. 


SECT,  xxix.]  EXCHANGE  OF  CURRENCIES.  147 

26.  At  $  13.625  per  cwt.,  what  cost  3cwt.  2qr.  Tib.  of 
sugar  ?  Ans.  $  48.53,9TV 

27.  At  $125.75  per  acre,  what  cost  37 A.  3R.  35rd.  ? 

Ans.  $  4774.57,0TV 

28.  At  $11.25  per  cwt.,  what  cost  17cwt.  2qr.  211b.  of 
rice?  Ans.  $198.98,4$. 

29.  What  cost  7J-  bales  of  cotton,  each  weighing  3.37cwt., 
at  $  9.37£  per  cwt.  ?  Ans.  $  244.85, 1TV 

30.  What  cost  7hhd.  49gal.  of  wine,  at  $  97.625  per  hogs- 
head  ?  Ans.  $  759.30,5f  % . 

31.  What  cost  7yd.  3qr.  3na.  of  cloth,  at  $4.75  per  yard  ? 

Ans.  $  37.70,3£. 

32.  What  cost  27T.  15cwt.  Iqr.  3£lb.  of  hemp,  at  $183.62 
per  ton  ?  Ans.  $  5098.03,7^. 

33.  What  is  the  cost  of  constructing  a  railroad  17m.  3fur. 
15rd.,  at  $1725.87,5  per  mile  ?  Ans.  $  30067.97 ,8££. 

34.  When  $624.53125  are  paid  for  17A.  3R.  15p.  of  land, 
what  is  the  cost  of  one  acre  ?  Ans.  $  35. 

35.  Paid  $494.53125  for  19T.   15cwt.  2qr.  141b.  of  hay; 
what  was  the  cost  per  ton  ?  Ans.  $  25. 

36.  How  much  land,  at  $40  per  acre,  can  be  obtained  for 
$1004.75  ?  Ans.  25A.  OR.  19p. 

37.  Bought  of  Queen  Victoria  9  acres  of  land,  for  which  I 
paid  157.753 125<£.     Required  the  price  per  acre. 

Ans.  17<£.  10s.  6£d. 

38.  If  $198.984375  are  paid  for  17cwt.  2qr.  211b.  of  rice, 
what  is  the  value  of  Icwt.  ?  Ans.  $11.25. 


SECTION  XXIX. 
EXCHANGE  OF  CURRENCIES. 

IT  is  well  known,  that,  in  different  States  of  the  Union,  the 
American  dollar  has  a  different  value  as  expressed  in  shillings 
and  pence.  The  origin  of  this  difference  is  thus  explained. 
Previous  to  the  formation  of  the  Constitution,  all  accounts  in 
this  country  were  kept  in  the  currency  of  Great  Britain,  and 
the  dollar  was  reckoned  at  4s.  6d.  sterling.  Owing,  however, 
to  the  want  of  money,  several  States  under  the  colonial  gov- 
ernment issued  Bills  of  Credit,  which  were  not  received  by  the 


148  EXCHANGE  OF  CURRENCIES.          [SECT.  ixix. 

British  merchants  in  payment  for  goods  at  their  par  value. 
Holders  of  those  bills  were  therefore  obliged  to  pay  a  larger 
nominal  amount  than  though  they  had  paid  in  sterling.  Thus 
eight  shillings  in  the  bills  of  New  York  passed  for  one  dollar,  or 
4s.  6d.  sterling.  In  the  bills  of  the  New  England  Colonies, 
where  the  depreciation  was  less,  six  shillings  made  a  dollar, 
and  in  South  Carolina  and  Georgia,  four  shillings  and  eight 
pence.  In  the  ordinary  reckonings  of  the  people,  shillings  and 
pence  are  still  considerably  used,  and  their  estimated  value  in 
different  States  is  as  follows. 

In  New  England,  Indiana,  Illinois,  Missouri,  Virginia,  Ken- 
tucky, Tennessee,  Mississippi,  Texas,  Alabama,  and  Florida, 
the  dollar  is  valued  at  6  shillings,  $  1  =  ^j£.  =  -^£. 

In  New  York,  Ohio,  and  Michigan,  the  dollar  is  valued  at 
8  shillings,  $  1  —  &£.  =  f  <£. 

In  New  Jersey,  Pennsylvania,  Delaware,  and  Maryland,  the 
dollar  is  considered  7  shillings  and  6  pence,  $  1  = 


In  North  Carolina  the  dollar  is  reckoned   at  10  shillings, 


In  South  Carolina  and  Georgia  4  shillings  8  pence  is  the 
value  of  a  dollar,  $  1  =  */&£.  =  &£. 

In  Canada  and  Nova  Scotia  the  dollar  is  valued  at  5  shillings, 


The  following  table  exhibits  the  legal  rates  of  interest  in 
the  United  States,  and  the  penalty  of  usury. 

STATES.     RATE  OF  INTEREST.        PENALTY  OF  USURY. 

Maine,  6  pr.  ct.  Forfeit  of  the  debt  or  claim. 

N.Hampshire,  6     "  Forfeit  of  threefold  the  usury. 

Vermont,  6     "  Recovery  in  an  action,  with  costs. 

Massachusetts,   6  Forfeit  of  threefold  the  usury. 

Rhode  Island,     6  Forfeit  of  the  usury  and  interest  on  the  debt. 

Connecticut,        6  Forfeit  of  the  whole  debt 

New  York,         7  Usurious  contracts  void. 

New  Jersey,       6  Forfeit  of  the  whole  debt. 

Pennsylvania,     6  Forfeit  of  the  whole  debt. 

Delaware,  6  Forfeit  of  the  whole  debt. 

Maryland,  6  On  tobacco  contracts,  8  per  ct.     Usurious 

contracts  void. 

Virginia,  6  Forfeit  double  the  usury. 

North  Carolina,  6  Contracts  for  usury  void.  Forfeit  double  the 

usury. 

South  Carolina,  7    "  Forfeit  of  interest  and  usury,  with  costs. 


SECT,  xxix.]          EXCHANGE  OF  CURRENCIES. 


149 


STATES. 

Georgia, 
Alabama, 


Louisiana, 

Tennessee, 

Kentucky, 

Ohio, 

Indiana, 

Illinois, 

Missouri, 

Michigan, 
Arkansas, 

Florida, 
Wisconsin, 

Iowa, 

Texas 


RATE  OF  INTEREST.       PENALTY  OF  USURY. 

8  per  ct.  Forfeit  of  three  times  the  usury,  and  con- 
tracts void. 

8     "        Forfeit  of  interest  and  usury. 
8     "       By  contract  as  high  as  10.     Recovery  in 

action  of  debt. 

5    "       Bank  6  ;  by  agreement  as  high  as  10;  con- 
tracts void. 

Contracts  void. 

Recovery  with  costs. 

Contracts  void. 

A  fine  of  double  the  usury. 

By  agreement  as  high  as  12.     Forfeit  of 
threefold  whole  amount  of  interest. 

By  agreement  as  high  as  10.    Forfeit  of  the 
interest  and  usury. 

Forfeit  of  the  usury  and  one  fourth  the  debt. 

By  agreement  as  high  as  10.     Usury  re- 
coverable ;  contracts  void. 

Forfeit  of  interest  and  usury. 

By  agreement  as  high  as  12. 
the  excess. 

By  agreement  as  high  as  12. 
the  excess. 


Forfeit  treble 
Forfeit  treble 


8 


Dist.  Columbia,  6     "       Contracts  void. 

NOTE.  —  On  debts  or  judgments  in  favor  of  the  United  States,  interest 
is  computed  at  the  rate  of  6  per  cent,  per  annum. 


In  order,  therefore,  to  change  any  of  the  above  currencies  to 
United  States  money,  the  shillings,  pence,  and  farthings,  if 
there  be  any,  must  first  be  reduced  to  decimals  of  a  pound, 
and  annexed  to  the  pounds. 


RULE.  —  Divide  the  pounds  by  the  value  of  a  dollar  in  the  given  cur- 
rency, EXPRESSED  BY  A  FRACTION  OF  A  POUND  ;  that  is,  to  change  the 
old  Neuo  England  currency  to  United  States  money,  divide  by  T30 ;  be- 
cause 6  shillings  is  T30  of  a  pound. 

To  change  the  old  currency  of  New  York,  <Sfc.,  to  United  States 
money,  divide  by  j0 ;  because  8  shillings  is  JQ  of  a  pound. 

To  change  the  old  currency  of  Pennsylvania,  <5fC.,  to  United  States 
money,  divide  by  §  ;  because  7  shillings  and  6  pence  is  $  of  a  pound. 

To  change  the  old  currency  of  South  Carolina  and  Georgia  to  United 
Slates  money,  divide  by  370  ;  because  4  shillings  and  8  pence  is  ^  of  a 
pound. 

To  change  Canada  and  Nova  Scotia  currency  to  United  States  money, 
;  because  5  shillings  is  %  of  a  pound. 

13* 


150  EXCHANGE  OF  CURRENCIES.          [SECT.  xxix. 

The  old  method  of  changing  English  sterling  money  to  United 
States  money  was,  to  divide  the  pounds  by  &,  and  the  quotient  was 
dollars ;  and,  to  change  dollars  into  English  sterling  money,  to  mul- 
tiply the  dollars  by  JQ,  and  the  product  was  pounds  sterling.  But, 
as  will  be  seen  by  a  note  on  page  151,  this  process  does  not  give  the 
present  value  of  a  pound  sterling. 

EXAMPLES. 

1.  Change  18«£.  4s.  6d.  of  the  old  New  England  currency 
to  United  States  money. 

18.225<£.  ~  A  =  $  60.75  Ans. 

In  this  example  we  reduce  the  4  shillings  and  6  pence  to  a 
decimal  of  a  pound,  which  we  find  to  be  .225.  This  decimal 
we  annex  to  the  pounds,  and  multiply  the  18.225<£.  by  10,  and 
divide  by  3,  and  it  produces  the  answer,  $  60.75.  The  reason 
for  this  process  has  already  been  shown. 

2.  Change  $  60.75  to  the  old  currency  of  New  England. 

$  60.75  X  A  =  18.225  =  18£.  4s.  6d.  Ans. 

The  decimal  .225  is  reduced  to  shillings  and  pence  by  Case 
IV.  of  Decimal  Fractions. 

3.  Change  7&£.  7s.  6d.  of  the  old  currency  of  New  Eng- 
land to  United  States  money.  Ans.  $  261.25. 

4.  Change  $  261.25  to  the  old  currency  of  New  England. 

Ans.  78o£.  7s.  6d. 

5.  Change  46<£.  16s.  6d.  of  the  old  currency  of  New  York 
to  United  States  money.  Ans.  $117.06£. 

6.  Change  $117.06£  to  the  old  currency  of  New  York. 

Ans.  46<£.  16s.  6d. 

7.  Change  387«£.  of  the  old  currency  of  Pennsylvania  to 
United  States  money.  Ans.  $1032. 

8.  Change   $1032   to   the   old   currency   of   Pennsylvania, 
Delaware,  and  Maryland.  Ans.  387<£. 

9.  Change  VZ£.  12s.  of  the  old  currency  of  South  Carolina 
and  Georgia  to  United  States  money.  Ans.  $  54. 

10.  Change  $  54  to  the  old  currency  of  South  Carolina  and 
Georgia.  Ans.  12<£.  12s. 

11.  Change  128<£.   18s.  6d.  of  Canada  and  Nova  Scotia  to 
United  States  money.  Ans.  $  515.70. 

12.  Change    $515.70    to    Canada   and    Nova    Scotia   cur- 
rency. Ans.  128<£.  18s.  6d. 


SECT,  xxix.]         EXCHANGE  OF  CURRENCIES.  151 

NOTE.  —  From  time  immemorial  $>1f  has  been  given  in  all  our  arithme- 
tics as  the  value  of  the  pound  sterling  in  United  States  Money.  It  is  time 
the  error  was  corrected. 

The  nominal  par  of  exchange  with  London,  as  expressed  in  reports  of 
exchange,  is  109.496-|->  or  very  nearly  1095,  being  9|  above  the  comput- 
ed par  of  $  4.444|,  represented  by  100. 

The  Bank  of  England  was  established  in  1694,  by  a  company  who  ad- 
vanced a  loan  of  £>  1,200,000  sterling  to  government.  Specie  payment  was 
suspended  in  1797,  and  resumed,  by  act  of  Parliament,  May  1,  1823. 

The  term  Sterling  is  derived  from  the  Eastcrli.ngs,  who  were  expert  re- 
finers from  the  eastern  part  of  Germany,  who  came  into  England  and  first 
established  the  standard  proportion  of  silver,  —  lloz.  2dwt.  fine  silver, 
and  18dwt.  alloy.  The  first  sterling  was  coined  in  1216.  In  the  reign 
of  Charles  the  Second  (1666)  a  new  gold  coinage  was  minted,  called  Guin- 
eas, from  the  country  from  which  the  gold  was  originally  brought.  In 
1816  (150  years  after)  the  guinea  was  superseded  by  a  new  coin,  called 
the  sovereign,  which  represents  the  pound  sterling.  The  guinea  (old  coin- 
age) weighs  129gfgr.,  standard.  The  sovereign  (new  coinage)  weighs 
123|pgr.,  standard.  The  standard  legal  fineness  of  gold  in  England  is 
22  carats,  or  f|$j.  The  standard  of  silver  is  lloz.  2dwt.  =  fg  =  $jfo 
The  sovereign  contains  precisely  HSggggr.  of  pure  gold. 

The  coinage  of  the  United  States  is  regulated  by  Congress.  By  the  last 
act  of  Congress,  January  18,  1837,  the  standard  for  both  gold  and  silver 
was  fixed  at  fggg,  • —  that  is,  suppose  any  of  our  gold  or  silver  coin  to  be  di- 
vided into  1000  equal  parts,  900  of  those  parts  are  pure  gold  or  silver,  and 
100  parts  are  alloy. 

By  this  act  the  eagle  weighs  258gr.  Troy,  standard. 
Containing,         .  .      •    232.2gr.  pure  gold, 

And       .         .  *       25.8gr.  alloy. 

Our  gold  coinage,  then,  is  2l|  carats  fine ;  our  silver  coinage  is  lOoz. 
16dwt.  fine,  —  6dwt.  short  of  sterling  fineness,  which.is  lloz.  2dwt.  The 
American  dollar  weighs  412£gr.,  standard,  containing  371  |gr.  pure  silver, 
and  414gr-  alloy.  The  alloy  in  our  gold  coins  is  mostly  silver,  and  in  our 
silver  coin  it  is  copper. 

The  ratio  of  gold  to  silver  in  our  coinage  is  15|g||  to  1,  —  that  is,  what- 
ever an  ounce  of  silver  may  be  worth,  an  ounce  of  gold  is  worth  15l§|| 
times  as  much. 

The  pound  sterling  under  the  above  act,  as  represented  by  the  sovereign 
of  legal  weight  and  fineness,  is  ^SIP  exactly,  =  $4.866+,  which  is 
the  real  gold  par  with  London. 

TO  REDUCE  STERLING  MONEY  TO  UNITED  STATES 
MONEY. 

RULE.  —  Express  the  shillings,  pence,  and  farthings  decimally;  then 
multiply  sterling  by  3?jS,  and  the  product  will  be  dollars,  <3fC. 

NOTE.  —  This  rule  supposes  the  sovereign,  which  represents  the  pound 

sterling,  to  be  f^p  fine,  and  to  contain  USslggr.  of  pure  gold.     But  the 
sovereign  falls  a  little  short  of  its  legal  weight  and  fineness.     So  that  its 


152  CIRCULATING  DECIMALS.  [SECT.  xxx. 

real  value  in  our  currency  does  not  vary  essentially  from  $4  84.  This  is 
the  value  assigned  to  it  by  act  of  Congress,  in  calculating  ad  valorem  duties 
in  our  custom-houses  on  goods  imported  from  England,  which  are  invoiced 
in  sterling  money.  Therefore,  multiply  sterling  by  $4.84  and  we  shall 
have  the  custom-house  and  market  par  value  of  sovereigns  or  pounds 
sterling. 

N.  B.  —  $4.444f  never  represented  the  true  value  of  the  pound  sterling 
in  the  United  States  currency. 

Under  the  act  of  Congress  of  the  2d  of  April,  1792,  establishing  the  mint 
and  regulating  the  coins  of  the  United  States,  the  value  of  the  pound  ster- 
ling was  $4  56-f- 

By  the  act  of  Congress  of  the  28th  of  June,  1834,  called  the  Gold,  Bill, 
the  value  of  the  pound  or  sovereign  was  $4  STj^Vs-  By  the  act  of  Con- 
gress of  the  18th  of  January,  1837,  supplementary  to  the  act  of  1834,  the 

value  of  the  pound  sterling  becomes  ft  TOKOS  4-  =  $4. 86723303— •  Sov- 
ereigns are  usually  valued  at  $4.85  at  the  banks. 


SECTION  XXX. 
INFINITE  OR  CIRCULATING  DECIMALS.* 

DEFINITIONS. 

1.  DECIMALS  produced  from  Vulgar  Fractions,  whose  denom- 
inators do  not  measure  their  numeratprs,  and  distinguished  by 
the  continual  repetition  of  the  sapie  f  gure  or  figures,  are  called 
infinite  or  circulating  decimals. 

2.  The  circulating  figures,  that  is,  those  that  are  continually 
repeated,  are  called  repetends.     If  only  the  same  figure  is  re- 
peated, it  is  called  a  single  repetend,  as  .11111  or  .5555,  and  is 
expressed  by  writing  the  figure  repeated  with  a  point  over  it. 
Thus  .  1 1 1 1 1  is  denoted  by  .  1 ,  and  .5555  by  .5. 

3.  If  the  same  figures  circulate  alternately,  it  is  called  a  com,' 
pound  repetend,  as  .475475475,  and  is  distinguished  by  putting 
a  point  over  the  first  and  last  repeating  figures ;  thus,  .475475475 
is  written  .475. 

4.  When  other  figures  arise  before  those  which  circulate,  it 
is  called  a  mixed  repetend  ;  as  .1246,  or  .17835. 

5.  Similar  repetends  begin  at  the  same  place  ;  as  .3  and  .6 ; 
or  5. 123  and  3.478. 

*  Infinite  or  circulating  decimals  being  less  important  for  use  than 
many  other  rules,  and  somewhat  difficult  in  their  operation,  the  student 
can  omit  them  until  he  reviews  the  Arithmetic. 


SECT,  xxx.]  CIRCULATING  DECIMALS.  153 

6.  Dissimilar  repetends  begin  at  different  places  ;  as  .986 
and  .4625. 

7.  Conterminous  repetends  end  at  the  same  place  ;  as  .631 
and  .465. 

8.  Similar  and  conterminous  repetends  begin  and  end  at  the 
same  place  ;  as  .1728  and  .4987. 

REDUCTION  OF  CIRCULATING  DECIMALS. 
CASE  I. 

To  reduce  a  simple  repetend  to  its  equivalent  vulgar  fraction. 

If  a  unit  with  ciphers  annexed  to  it  be  divided  by  9  ad  infini- 
tum,  the  quotient  will  be  one  continually  ;  that  is,  if  ^  be  re- 
duced to  a  decimal,  it  will  produce  the  circulate  .1  ;  and  since 
.1  is  the  decimal  equivalent  to  £,  .2  will  be  equivalent  to  f ,  .3  to 
f ,  and  so  on,  till  .9  is  equal  to  f  or  1.  Therefore  every  single 
repetend  is  equal  to  a  vulgar  fraction,  whose  numerator  is  the 
repeating  figure,  and  denominator  9.  Again,  ^,  or  ^^,  being 
reduced  to  decimals,  makes  .01010101,  and  .001001001  ad  in- 
Jinitum,  —  .01  and  .001  ;  that  is,  FV  =  .61,  and  fffa  =  .001  ; 
consequently  ¥2F  3=  .02,  and  ^f  F  =  .002  ;  and,  as  the  same  will 
hold  universally,  we  deduce  the  following 

RULE.  —  Make  the  given  decimal  the  numerator,  and  let  the  denomi- 
nator be  a  number  consisting  of  as  many  nines  as  there  are  recurring 
places  in  the  repetend. 

If  there  be  integral  figures  in  the  circulate,  as  many  ciphers  must  be 
annexed  to  the  numerator  as  the  highest  place  of  the  repetend  is  distant 
from  the  decimal  point. 

EXAMPLES. 

1.  Required  the  least  vulgar  fraction  equal  to  .6  and  .123. 

.6  =  f  =  f  Ans.  .123  =  iff  =  #&  Ans. 

2.  Reduce  .3  to  its  equivalent  vulgar  fraction.          Ans.  £. 

3.  Reduce  1.62  to  its  equivalent  vulgar  fraction.  Ans.  lf£. 

4.  Reduce  .769230  to  its  equivalent  vulgar  fraction. 

Ans.  if. 
CASE  II. 

To  reduce  a  mixed  repetend  to  its  equivalent  vulgar  fraction. 


154  CIRCULATING  DECIMALS.  [SECT.  XM. 

1.  What  vulgar  fraction  is  equivalent  to  .138  ? 

OPERATION. 

.138  =  TVS,  +  yfo  =  iU  +  irfcr  =  «*  =  A  Ans. 
As  this  is  a  mixed  circulate,  we  divide  it  into  its  finite  and 
circulating  parts  ;  thus  .138  =  .13,  the  finite  part,  and  .008  the 
repetend  or  circulating  part  ;  but  .13  =  T\y^  ;  and  .008  would 
be  equal  to  f,  if  the  circulate  began  immediately  after  the 
place  of  units  ;  but,  as  it  begins  after  the  place  of  hundreds,  it  is 

f  of  T*IT  =  *»*•      Therefore  .138  =  Jfo  +  ^  =  $tf  + 
Q.  E.D. 


RULE.  —  To  as  many  nines  as  there  are  figures  in  the  repetend,  an- 
nex as  many  ciphers  as  there  are  finite  places  for  a  denominator  ;  mul- 
tiply the  nines  in  the  denominator  by  the  finite  part,  and  add  the  repeat- 
ing decimal  to  the  product  for  the  numerator.  If  the  repetend  begins 
in  some  integral  place,  the  finite  value  of  the  circulating  must  be  added 
to  the  finite  part. 

2.  What  is  the  least  vulgar  fraction  equivalent  to  .53  ? 

Ans.  T^-. 

3.  What  is  the  least  vulgar  fraction  equivalent  to  .5925  ? 

Ans.  £f  . 

4.  What  is  the  least  vulgar  fraction  equivalent  to  .008497133  ? 

Ans. 


5.  What  is  the  finite  number  equivalent  to  3.62  ? 

Ans.  31Jf 

CASE  III. 

To  make  any  number  of  dissimilar  repetends  similar  and 
conterminous. 

1.  Dissimilar  made  similar  and  conterminous. 

OPERATION.  Any    given    repetend    whatever, 

9.167  =    9.61767676     whether  single,  compound,  pure,  or 

14  g      __  14  60000000     mixe(^  mav  De  transformed  into  an- 

other repetend,  that  shall  consist  of 

:    3.16555555     an  equal  or  greater  number  of  figures 

12.432  =  12.43243243     at  pleasure  .  thus  4  may  ^  chan?0(] 

8.181—    8.18181818     into  ,44  or  .444  ;  and  .29  into   - 

1.307  =    1.30730730    or  ^^      And  as  some  of  the  circu. 

lates  in  this  question  consist  of  one,  some  of  two,  and  others  of 
three  places  ;  and  as  the  least  common  multiple  of  1,  2,  and  3 


SECT,  xxx.]  CIRCULATING   DECIMALS.  155 

is  6,  we  know  that  the  new  repetend  will  consist  of  6  places, 
and  will  hegin  just  so  far  from  unity  as  is  the  farthest  among 
the  dissimilar  repetends,  which,  in  the  present  example,  is  the 
third  place. 

RULE.  —  Change  the  given  repetends  into  other  repetends ,  which  shall 
consist  of  as  many  figures  as  the  least  common  multiple  of  the  several 
number  of  places  found  in  all  the  repetends  contains  units. 

2.  Make  3.671,  1. 007 i,  8.52,  and  7.616325  similar  and  con- 
termio  us. 

3.  Make  1.52,  8.7156,  3.567,  arid  1.378  similar  and  conter- 
minous. 

4.  Make  .0007,.  14 14 14, and  887.1  similar  and  conterminous. 

CASE  IV. 

To  find  whether  the  decimal  fraction  equal  to  a  given  vulgar 
fraction  be  finite  or  infinite,  and  of  how  many  places  the  repe- 
tend will  consist. 

RULE.  —  Reduce  the  given  fraction  to  its  least  terms,  and  divide  the 
denominator  by  2,  5,  or  10,  as  often  as  possible.  If  the  whole  denomi- 
nator vanish  in  dividing  by  2,  5,  or  10,  the  decimal  will  be  finite,  and 
will  consist  of  so  many  places  as  you  perform  divisions.  If  it  do  not 
vanish,  divide  9999,  cf-c.,  by  the  result  till  nothing  remain,  and  the  num- 
ber of  9 "s  used  will  show  the  number  of  places  in  the  repetend;  which 
will  begin  after  so  many  places  of  figures  as  there  are  10 '5,  2's,  or  5's 
used  in  dividing. 

NOTE.  —  In  dividing  1.0000,  &c.,  by  any  prime  number  whatever,  except 
2  or  5,  the  quotient  will  begin  to  repeat  as  soon  as  the  remainder  is  1.  And 
since  9999,  &c.,  is  less  than  10000,  &c.,  by  1,  therefore  9999,  &c.,  divid- 
ed by  any  number  whatever,  will  leave  a  0  for  a  remainder,  when  the  re- 
peating figures  are  at  their  period.  Now  whatever  number  of  repeating 
figures  we  have  when  the  dividend  is  1,  there  will  be  exactly  the  same 
number  when  the  dividend  is  any  other  number  whatever.  For  the  prod- 
uct of  any  circulating  number  by  any  other  given  number  will  consist  of 
the  same  number  of  repeating  figures  as  before.  Thus,  let  378137813781, 
&c.,  be  a  circulate,  whose  repeating  part  is  3781.  Now  every  repetend 
(3781),  being  equally  multiplied,  must  produce  the  same  product.  For 
these  products  will  consist  of  more  places,  yet  the  overplus  in  each,  being 
alike,  will  be  carried  to  the  next,  by  which  means  each  product  will  be 
equally  increased,  and  consequently  every  four  places  will  continue  alike. 
And  the  same  will  hold  for  any  other  number  whatever.  Hence  it  ap- 
pears, that  the  dividend  may  be  altered  at  pleasure,  and  the  number  of 
places  in  the  repetend  will  be  still  the  same  ;  thus,  n  =  .09,  and  f*T  =  .27, 
where  the  number  of  places  in  each  are  alike  ;  and  the  same  will  be  true 
in  all  cases. 


156  CIRCULATING  DECIMALS.  [SECT.  MXI. 

EXAMPLES. 

1.  Required  to  find  whether  the  decimal  equal  to  -f^f  be 
finite  or  infinite  ;  and  if  infinite,  of  how  many  places  the  repe- 
tend  will  consist. 

(2)        (2)        (2) 

fffi  =  2)T%  =  8  =  4  =  2=1;  therefore,  because  the  de- 
nominator vanishes  in  dividing,  the  decimal  is  finite,  and  con- 
sists of  four  places  ;  thus,  16)3;${$$. 

2.  Required  to  find  whether  the  decimal  equal  to  //<&  be 
finite  or  infinite  ;  and,  if  infinite,  of  how  many  places  that  rep- 
etend  will  consist. 

JM,  =  -&  2)112  =  56  =  28  =  14  =  7.  Thus,  ''ffff  Jf ; 
therefore,  because  the  denominator,  112,  did  not  vanish  in  di- 
viding by  2,  the  decimal  is  infinite  ;  and  as  six  9's  were  used, 
the  circulate  consists  of  six  places,  beginning  at  the  fifth  place, 
because  four  2's  were  used  in  dividing. 

3.  Let  TV  be  the  fraction  proposed. 

4.  Let  ^4j-  be  the  fraction  proposed. 


SECTION  XXXI. 
ADDITION  OF  CIRCULATING  DECIMALS. 

EXAMPLE. 

1.  Let  3.5  +  7.651  +  1.765  +  6.173  +  51.7  +  3.7+27.631 
and  1.003  be  added  together. 

OPERATION, 

Dissimilar.    Similar  and  Conterminoui. 

3.5      =    3.5555555 

rj  -gg'j 7  6516516  Having  made  all  the  numbers  similar 

".    .                          .  and  conterminous  by  Sect.  XXX.,  Case 

1.765  =    1.7657657  ni.,  we  add  the  first  six  columns,  as  in 

6.173r=    6.1737373  Simple  Addition,  and  find  the  sum  to 

51.7      =  51.7777777  be   3591224  =  W?Ws  =  3.591227. 

3.7      =    3.7000000  The   repeating   decimals  .591227  we 

27.631  =  27.6316316  write  in  their  Pr°Per  place,  and  carry  3 

-  •Q~'  _  nAonrvnA  to  me  next  column,  and  then  proceed 

as  in  whole  numbers. 
103.2591227 


«ECT.  xxxn.]  CIRCULATING  DECIMALS.  157 

RULE.  —  Make  the  repetends  similar  and  conterminous,  and  find  their 
sum,  as  in  common  Addition.  Divide  this  sum  by  as  many  9's  as 
there  are  places  in  the  repetend,  and  the  remainder  is  the  repetend  of  the 
sum,  which  must  be  set  under  the  figures  added,  with  ciphers  on  the  left 
when  it  has  not  so  many  places  as  the  repetends.  Carry  the  quotient  of 
this  division  to  the  next  column,  and  proceed  with  the  rest  as  with  finite 
decimals. 

2.  Add  27.56  +  5.632  +  6.7  +  16.356  +  .71  and  6.1234 
together.  Ans.  63.1690670868888. 

3.  Add  2.765  +  7.16674  +  3.671  +  .7  and  .1728  together. 

Ans.  14.55436. 

4.  Add  5.16345  +  8.6381  +  3.75  together. 

Ans.  17.55919120847374090302. 

5.  Reduce  the  following  numbers  to  decimals,  and  find  their 
sum  :  £,  f ,  and  £.  Ans.  .587301. 


SECTION  XXXII. 
SUBTRACTION  OF  CIRCULATING  DECIMALS. 

EXAMPLE. 

1.  From  87.1645  take  19.479167. 

OPERATION.  Having  made  the  numbers  similar 

87.164*5      —  87.164545     and  conterminous,  we  subtract  as  in 
IQ  A^Q-ia^f  —  10.  AiQifvi     whole  numbers,  and  find  the  remain- 

Xc'.i±/l7  J.O/    —    lU.'db/cHO  /         ,  -    ,  .         ',  ,        cor-ro    c 

— :     der  of  the  circulate  to  be  5378,  from 

67.685377     which  we  subtract  1,  and  write  the 
remainder  in  its  place,  and  proceed 

with  the  other  part  of  the  question  as  in  whole  numbers.     The 

reason  why  1  should  be  added  to  the  repetend  may  be  shown  as 

OPERATION    f°tt°ws-    The  minuend  may  be  considered  16f|ff, 

an(^  ^e  subtrahend  7f£f|- ;  we  then  proceed  with 

79  if  t      these  numbers  as  in  Case  II.  of  Subtraction  of  Vulgar 

[fit*      Fractions  ;  and  the  numerator  5377  will  be  the  re- 

8J-fi&      peating  decimal.     Q.  E.  D. 

RULE.  —  Make  the  repetends  similar  and  conterminous,  and  subtract 
as  usual ;  observing,  that  if  the  repetend  of  the  subtrahend  be  greater 
than  the  repetend  of  the  minuend,  then  the  remainder  on  the  right  must 
be  less  by  unity  than  it  would  be  if  tJie  expressions  were  finite. 
14 


156  CIRCULATING  DECIMALS.  [SECT.  xxxm. 

2.  From  7.1  take  5.t)2.  Ans.  2.08. 

3.  From  315.87  take  78.0378.    Ans.  237.838072095497. 

4.  Subtract  |  from  f .  Ans.  .079365. 

5.  From  16.1347  take  11.0884.  Ans.  5.0462. 

6.  From  18.1678  take  3.27.  Ans.  14.8951. 

7.  From  3.123  take  0.71.  Ans.  2.405951*. 

8.  From  £  take  f£  Ans.  .246753. 

9.  From  f  take  f.  Ans.  .158730. 

10.  From  T9T  take  TV  Ans.  .1764705882352941. 

11.  From  5.12345  take  2.3523456. 

Ans.  2.77 1 105582 1666927777988888599994. 


SECTION  XXXIII. 
MULTIPLICATION  OF  CIRCULATING  DECIMALS. 

1.  Multiply  .36  by  .25. 

First  Method.  In  the  first  method, 

OPERATION.  we  reduce  the  num- 

ofi_36_—  4.  9^—2  _i     5   —23      bers   to   vulgar    frac- 
•36    -  if  —  A,  -25  -  TIT  +  w  -  w     tiongj  and  tlfen  muki. 

A  X  f  £  =  F9^  =  .0929  Answer.  ply  and  reduce  them 

2.  Multiply  582.347  by  .08. 

Second  Method.  In  the  second  method, 

OPERATION.  we  multiply  as  in  whole 

582.1347  X  .08  =  46.58778  Answer,  numbers,  but  we  add  two 

units  to  the  product ;  for 

8  x  347  =  2776  =  %¥  =  2Ht-     Thus  we  see  the  repeating 
number  is  778. 

RULE.  —  Turn  both  the  terms  into  their  equivalent  vulgar  fractions, 
and  find  the  product  of  those  fractions  as  usual.  Then  change  the  vul- 
gar fraction  expressing  the  product  into  an  equivalent  decimal,  and  it 
will  be  the  product  required.  But,  if  the  multiplicand  ONLY  has  a  rep- 
etend,  multiply  as  in  whole  numbers,  and  add  to  tfte  right-hand  place  of 
the  product  as  many  units  as  there  are  tens  in  the  product  of  the  left- 
hand  place  of  the  repetend.  Tlie  product  will  then  contain  a  repetend 
whose  places  are  equal  to  those  in  the  multiplicand. 


BBCT.  xxxv.]  MISCELLANEOUS.  159 

3.  Multiply  87.32586  by  4.37.  Ans.  381.6140338. 

4.  Multiply  3. 145  by  4.297.  Ans.  13.5169533. 

5.  What  is  the  value  of  .285714  of  a  guinea  ?       Ans.  8s. 

6.  What  is  the  value  of  .461607142857  of  a  ton  ? 

Ans.  9cwt.  Oqr.  261b. 

7.  What  is  the  value  of  .284931506  of  a  year  ? 

Ans.  104da. 


SECTION  XXXIV. 
DIVISION  OF  CIRCULATING  DECIMALS. 

1.  Divide  .54  by  .15. 

OPERATION. 

.54  —  |*  —  6  Having  reduced  the  num- 

1  k  _   _j     I     B     _i4_     7         t>ers  to  vulgar  fractions,  we 

e  T  JL^Jr  >7v  =Tw  divide  °ne  by  the  °ther'  and 
TT  '  ¥?  l  r  .  T  ~T  TT  *  change  the  quotient  to  a  de- 
*?f  =  3ff  =  3.506493  Ans.  cimaj; 

RULE.  —  Change  both  the  divisor  and  the  dividend  into  their  equiva- 
lent vulgar  fractions,  and  find  their  quotient  as  usual.  Change  the  vul- 
gar fraction  expressing  the  quotient  into  its  equivalent  decimal,  and  it 
will  be  the  quotient  required. 

2.  Divide  345.8  by  .6.  Ans.  518.83. 

3.  Divide  234.6  by  .7.  Ans.  301.714285. 

4.  Divide  .36  by  .25.  Ans.  1.4229249011857707509881. 


SECTION  XXXV. 
MENTAL  OPERATIONS  IN  FRACTIONS,  &c. 

IF  any  number  be  divided  into  two  equal  parts,  and  into 
two  unequal  parts,  the  product  of  the  two  unequal  parts  togeth- 
er with  the  square  of  half  the  difference  of  the  two  unequal 
parts  is  equal  to  the  square  of  one  of  the  equal  parts.  Also, 

The  product  of  any  two  numbers  is  equal  to  the  square  of 


160  MISCELLANEOUS.  [SECT.  xxxr. 

half  their  sum,  less  the  square  of  half  their  difference.     See 
Euclid's  Elements,  Book  Second,  Proposition  Fifth. 

NOTE.  —  A  number  is  said  to  be  squared  when  it  is  multiplied  by  itself; 
thus,  the  square  of  5  is  5  X  5  =  25. 

From  the  above  proposition  we  deduce  the  following  rules. 
To  multiply  any  number  containing  a  half  by  itself. 

RULE  1.  —  Multiply  the  whole  number  given  in  the  question  by  the 
next  larger  whole  number,  and  to  the  product  add  the  square  of  the  half 

1.  Multiply  5£  by  5J. 

OPERATION. 

:=30    Ans. 


NOTE.  —  The  whole  number  given  is  5,  and  the  next  larger  whole  num- 
ber is  6. 

2.  Multiply  7£  by  7£.  Ans.  56  J. 

3.  Multiply  3|  by  3£.  Ans.  Iftf, 

4.  Multiply  9£  by  9£.  Ans.  90,  . 

5.  Multiply  11£  by  11£.  Ans.  132£. 

6.  Multiply  2(4  by  20£.  Ans.  420T. 

7.  Multiply  30|  by  30£.  Ans.  930,  . 

le  will  hold  good  if  we  multiply  any  num- 


NOTE  —  The  same  principl 
ber  by  itself  whose  unit  is  a  5. 


RULE  2.  —  Take  the  next  least  number  that  ends  in  a  cipher,  and  mul- 
tiply it  by  the  next  larger  number  ending  in  a  cipher,  and  add  to  the 
product  the  square  of  5  =  25,  and  the  result  will  be  the  product. 

8.  Multiply  25  by  25.  Ans.  625. 
The  next  less  number  ending  in  a  cipher  is  20,  and  the  next 

larger  is  30;  30x20  =  600;  5x5  =  25;  600  +  25  =  625 
Ans. 

9.  Multiply  35  by  35.  Ans.  1225. 

10.  Multiply  85  by  85.  Ans.  7225. 

11.  Multiply  95  by  95.  Ans.  9025. 

To  find  the  product  of  two  mixed  numbers,  whose  fractional 
part  is  a  half,  and  whose  difference  is  a  unit. 

RULE  3.  —  Multiply  the  larger  number  idthout  the  fraction  by  itself, 
and  from  the  product  subtract  the  fractional  part  multiplied  by  itself,  and 
the  result  will  be  tfte  product. 

12.  Multiply  6£  by  7f  Ans.  48J. 


SECT,  xxxv.]  MISCELLANEOUS.  161 

OPERATION. 

7  X  7  =  49  ;  £  x  £  —  £  ;  49  —  £  =  48}  Ans. 

13.  Multiply  8£  by  9£.  Ans.  80}. 

14.  Multiply  11£  by  12J.  Ans.  143f. 

15.  Multiply  19^-  by  20£.  Ans.  399£. 

16.  Multiply  89i  by  90  J.  Ans.  8099}. 

NOTE.  —  If  the  fractional  parts  of  the  numbers  approach  within  a  J,  £,  or 
|,  &c.,  of  the  larger  number,  the  principle  is  the  same. 

17.  What  is  the  product  of  4f  multiplied  by  5£.  Ans.  24f . 

OPERATION. 

5X5  =  25;  ix£  =  i;  25  -  J  =  24|  Ans. 

18.  Multiply  7}  by  8±.  Ans.  63|f . 

19.  Multiply  9$  by  lOf  Ans.  99ff . 

20.  Multiply  8|  by  9f .  Ans.  80^. 

21.  Multiply  19^  by  20TV  Ans.  399^. 

To  find  the  product  of  two  numbers,  one  of  which  is  as  much 
less  than  either  20,  30,  40,  &c.,  as  the  other  is  more  than  either 
of  these  numbers. 

RULE  4.  —  Multiply  the  20,  30,  or  40,  as  the  case  may  be,  by  itself, 
and  subtract  from  the  product  the  square  of  half  the  difference  of  the 
two  numbers  to  be  multiplied,  and  the  result  will  be  the  product. 

22.  Multiply  28  by  32.  Ans.  896. 
We  find  that  28  is  as  much  less  than  30  as  32  is  more  than 

30 ;  we  therefore  multiply  30  by  30  =  900,  and  from  this  prod- 
uct we  subtract  the  square  of  2  =  4  ;  900  —  4  =  896  Ans. 

23.  What  is  the  product  of  75  by  85  ?  Ans.  6375. 

24.  What  is  the  product  of  83  by  77  ?  Ans.  6391. 

25.  What  is  the  product  of  97  by  103  ?  Ans.  9991. 

26.  What  is  the  product  of  17  by  23  ?  Ans.  391. 

27.  What  cost  18cwt.  of  steel,  at  $22  per  cwt.  ? 

Ans.  $  396. 

28.  What  cost  27  tons  of  hay,  at  $  33  per  ton  ? 

Ans.  $891. 

29.  What  cost  64  gallons  of  oil,  at  $  0.56  per  gallon  ? 

Ans.  $35.84. 

30.  What  cost  28  tons  of  hay,  at  $  32  per  ton  ? 

Ans.  $  886. 

31.  What  cost  49  tons  of  iron,  at  $  51  per  ton  ? 

Ans.  $2499. 
14* 


162  MISCELLANEOUS.  [SECT,  MXVI. 

SECTION  XXXVI. 

QUESTIONS  TO  BE  PERFORMED  BY  ANALYSIS. 

- 

I.  If  a  man  travel  48  miles  in  12  hours,  how  far  will  he 
travel  in  17  hours  ?  Ans.  68  miles. 

[The  following  is  the  most  obvious  solution  of  this  question.  If  he 
travel  48  miles  in  12  hours,  in  ]  hour  he  will  travel  1*5  of  48  miles, 
which  is  4  miles.  Then  if  be  travel  4  miles  in  1  hour,  he  will  in  17  hours 
travel  17  times  as  far,  which  is  68  miles,  the  answer.] 

2.  If  72  pounds  of  beef  cost  $  6.48,  what  will  1  pound  cost  ? 
what  will  675  pounds  cost  ?  Ans.  $  60.75. 

3.  If  £  of  a  dollar  buy  1  pound  of  sugar,  how  much  may  be 
bought  for  $  1  ?  how  much  for  $  29 1  ?  Ans.  2391b. 

4.  If  a  hogshead  of  wine  cost  $  73.50,  what  is  the  value  of 
1  gallon  ?  what  cost  17hhd.  45  gallons  ?         Ans.  $  1302.00. 

5.  Bought  11  bushels  of  rye  for  $  9.00  ;  what  cost  25  bush- 
els  ?  Ans.  $  20.45^T. 

6.  If  a  crew  of  15  hands  consume  in  3  months  1620  pounds 
of  beef,  how  much  would  be  sufficient  for  27  hands  for  the 
same  time  ?  Ans.  29161b. 

7.  Bought  9  yards  of  flannel  for  $  7.00 ;  what  would  be  the 
value  of  37|£  yards  ?  Ans.  $  29. 3  If 

8.  If  a  certain  field  will  pasture  8  horses  nine  weeks,  how 
long  will  it  pasture  23  horses  ?  Ans.  3^  weeks. 

9.  If  7T32-  dozen  of  hats  cost  $  318.50,  what  will  19|£  doz- 
en cost  ?  Ans.  $  874.95f  f 

10.  If  1  ton  of  hay  cost  $  25.00,  what  will  17  tons  13cwt.  19 
pounds  cost  ?  Ans.  $  441.46-J-. 

II.  What  part  of  9f  is  25  ?  Ans.  2ff. 

12.  What  part  of  25  is  9£  ?  Ans.  T^0. 

13.  If  $  25  will  pay  for  7-jj-  yards  of  broadcloth,  what  would 
be  the  price  of  97  yards  ?  Ans.  $  328.8 If  £. 

14.  If  $  47.25  will  pay  for  the  keeping  of  7  horses  2  months, 
what  would  it  cost  to  keep  43  horses  for  the  same  time  ? 

Ans.  $290.25. 

15.  If  7£  yards  of  cloth  a  yard  wide  is  sufficient  to  make  a 
cloak,  how  many  yards  would  it  take  if  the  cloth  was  If  yards 
wide  ?  Ans.  4f  yards. 

16.  If  a  barrel  of  beer  will  last  10  men  a  week,  how  long 
would  it  last  1  man  ?  how  long  37  men  ?         Ans.  1§^  days. 

17.  If  5  calves  are  worth  9  sheep,  how  many  calves  will 
purchase  108  sheep  ?  Ans.  60  calves. 


SECT,  xxxvi.]  MISCELLANEOUS.  163 

18.  If  11  yards  of  cotton  3  quarters  wide  are  sufficient  to 
line  a  garment,  how  many  yards  would  it  require  that  were  5 
quarters  wide  ?  .     Ans.  6f  yards. 

19.  If  7  pairs  of  shoes  will  purchase  2  pairs  of  boots,  how 
many  pairs  of  shoes  would  it  take  to  buy  18  pairs  of  boots  ? 

Ans.  63  pairs. 

20.  If  four  gallons  of  vinegar  be  worth  7  gallons  of  cider, 
what  quantity  of  vinegar  would  it  take  to  buy  47  gallons  of 
cider  ?  Ans.  26f  gallons. 

21.  If  a  man  travel  377  miles  in  15  days,  how  far  would  he 
travel  in  1  day  ?  how  far  in  100  days  ?      Ans.  2513£  miles. 

22.  If  12  men  can  dig  a  ditch  50  feet  long,  4  feet  wide,and 
3  feet  deep,  in«30  days,  how  long  would  it  take  1  man  ?  how 
long  47  men  ?  Ans.  7f  \  days. 

23.  If  5  barrels  of  flour  cost  $25.75,  what  will  39  barrels 
cost  ?  Ans.  $  200.85. 

24.  Bought  17  acres  of  land  for  $791.01  ;  what  would  98 
acres  3  roods  and  14  perches  cost  ?  Ans.  $4598.90,8£. 

25.  Bought  97f  yards  of  cloth  for  $  275.20 ;  what  is  the 
price  of  7  yards  ?  Ans.  $  19.78-f . 

26.  If  a  pole  7  feet  long  cast  a  shadow  of  5  feet,  how  high 
is  that  steeple  whose  shadow  is  97  feet?          Ans.  135f  feet. 

27.  Gave  3£cwt.  of  sugar,  at  $  9,  for  -f^f  of  an  acre  of  land  ; 
how  much  sugar  would  it  have  required  to  purchase  an  acre  ? 

Ans.  5f  f  cwt. 

28.  If  a  vessel  sail  47f  miles  in  3f  hours,  how  far  would  it 
sail  in  1  week  ?  Ans.  2425211Jm. 

29.  James  can  mow  a  field  in  7  days,  by  laboring  10  hours  a 
day ;  how  many  days  would  it  take  him  to  perform  the  work  by 
laboring  12  hours  a  day  ?  Ans.  5|  days. 

30.  If  -jZj-  of  a  lot  of  land  is  worth  $42.12,  what  is  the  value 
of  f  of  it?  Ans.  $29.4 If 

31.  If  a  man  can  earn  $  10.27  in  f  of  a  week,  how  much 
would  he  earn  in  a  month  ?  Ans.  $71.89. 

32.  If  18  men  can  reap  72  acres  in  5  days,  how  long  would 
it  take  6  men  to  perform  the  labor  ?  Ans.  15  days. 

33.  If  19  gallons  of  wine  can  be  bought  for  $  25,  how  many 
gallons  will  $  71.25  buy  ?  Ans.  54/g-  gallons. 

34.  If  a  penny  loaf  weighs  8  ounces  when  wheat  is  $  1  a 
bushel,  what  should  it  weigh  when  wheat  is  sold  for  $  1.25  a 
bushel  ?  Ans.  6f  ounces. 

35.  If  a  basket  which  contains  1£  bushels  must  be  filled  with 
apples  7  times  to  make  one  barrel  of  cider,  how  many  barrels 
may  be  made  by  its  being  filled  126  times  ?  Ans.  18bbl. 


164  SIMPLE  INTEREST.  [SECT.  XXXTII. 

36.  If  a  cloak  can  be  made  of  4|  yards  of  cloth  that  is  If 
yards  wide,  how  many  yards  would  it  take  of  cloth  that  is  £  of 
a  yard  wide  ?  Ans.  6yd.  Iqr.  3f  na. 

37.  If  a  box  4  feet  long,  2  feet  wide,  l£  feet  high,  contains 
300  pounds  of  sugar,  how  much  will  a  box  that  is  8  feet  long,  4 
feet  wide,  and  3  feet  high,  contain  ?  Ans.  24001b. 

38.  How  much  in  length,  that  is  12f  rods  in  breadth,  will 
make  an  acre  ?  Ans.  12y8Trd. 

39.  Sound,  uninterrupted,  moves  1 142  feet  in  a  second  ;  how 
long,  after  a  cannon's  being  discharged  at  Boston,  is  the  time 
before  it  is  heard  at  Bradford,  the  nearest  distance  being  25^ 
miles  ?  Ans.  2  minutes. 

40.  Bought  ^T  of  a  ton  of  potash,  and  sold  -f£  of  it  for 
$  46.70  ;  what  was  the  value  of  a  ton  ?  Ans.  $  93.40. 

41.  If  -j4T  of  a  lot  of  land  be  worth  $97,  what  would  the 
whole  lot  be  worth  ?  Ans.  $  266.75. 

42.  If  19  pounds  of  salmon  be  worth  50  pounds  of  beef,  how 
much  salmon  would  buy  77  pounds  of  beef?        Ans. 

43.  What  part  of  17T\  is  4£  ?  Ans. 

44.  What  part  of  lls.  3d.  is  15s.  ?  Ans.  |. 

45.  What  part  of  7±  yards  is  3|  ells  English  ?     Ans. 


SECTION  XXXVII. 
SIMPLE   INTEREST. 

INTEREST  is  the  compensation  which  the  borrower  of  money 
makes  to  the  lender. 

PRINCIPAL  is  the  sum  lent. 

AMOUNT  is  the  interest  added  to  the  principal. 

PER  CENT.,  a  contraction  of  per  centum ,  is  the  rate  establish- 
ed by  law,  or  that  which  is  agreed  on  by  the  parties,  and  is  so 
much  for  a  hundred  dollars  for  one  year. 

CASE  I. 

GENERAL  RULE.  —  Let  the  per  cent,  be  considered  as  a  decimal  of  a 
hundred  dollars,  and  multiply  the  principal  by  it,  and  the  product  is  the 
interest  for  one  year.  But,  if  it  be  required  to  find  the  interest  for  more 
than  one  year,  multiply  the  product  by  the  number  of  years. 

NOTE.  —  The  decimal  for  6  per  cent,  is  06 ;  for  7  per  cent.,  .07  ;  for  8 
per  cent.,  .08  ;  for  9$  per  cent.,  .0925  ;  for  2£  per  cent.,  .025,  &c.  The 
decimal  must  be  pointed  off  as  in  Multiplication  of  Decimal  Fractions. 


SECT,  xxxvii.]  SIMPLE  INTEREST.  165 

This  rule  is  obvious  from  the  fact  that  the  rate  per  cent,  is 
such  a  part  of  every  hundred  dollars.  Thus  6  per  cent,  is  T£F 
of  the  principal. 

NOTE.  —  When  no  particular  per  cent,  is  named,  6  per  cent,  is  to  be 
understood,  as  it  is  the  legal  interest  in  the  New  England  States.  See 
page  148. 

1.  What  is  the  interest  of  8  144  for  1  year  ?    Ans.  $  8.64. 

OPERATION. 

Qg  There  being  two  places  of  decimals  in  the 

— —   .          multiplier,  we  point  off  two  in  the  product. 
8  8.64  Ans. 

2.  What  is  the  interest  of  $  78.78  for  3  years  ? 

OPERATION. 

$78.78 

.06         There  being  two  places  of  decimals  in  the 

4  72  68  multiplicand,  and  two  in  the  multiplier,  we  point 

3  off  four  places  in  the  product. 

$  14. 18,04  Ans. 

NOTE.  —  It  is  a  custom  with  merchants  to  reject  mills  in  their  compu- 
tations, but  when  the  decimal  of  a  cent  exceeds  5  they  add  1  to  the  num- 
ber of  cents.  Thus,  they  would  reckon  $81.93,8  to  be  in  value  $  81.94. 

3.  What  is  the  interest  of  8  675  for  1  year  ?  Ans.  8  40.50. 


4.  What  is  the  interest  of  8  1728  for  1  year  ? 

Ans.  8  103.68. 

5.  What  is  the  interest  of  $  19.64  for  2  years  ? 

Ans.  $  2.35,6. 

6.  What  is  the  interest  of  8  896.28  for  3  years  ? 

Ans.  8  161.33. 

7.  What  is  the  interest  of  $  349.25  for  10  years  ? 

Ans.  8  209.55. 

8.  What  is  the  interest  of  8  3967.87  for  2  years  ? 

Ans.  $476.14,4. 

9.  What  is  the  interest  of  8  123.45  for  6  years  ? 

Ans.  $  44.44,2. 

10.  What  is  the  interest  of  $  89.25  for  50  years  ? 

Ans.  $267.75. 

11.  What  is  the  interest  of  $  17.25  for  7  years  ? 

Ans.  $7.24,5. 

12.  What  is  the  interest  of  $  29.19  for  9  years  ? 

Ans.  8  15.76,2. 

13.  What  is  the  interest  of  8  617.56  for  25  years  ? 

Ans.  8  926.34. 


166  SIMPLE   INTEREST.  [SECT,  xxxvn. 

14.  What  is  the  amount  of  $  31.75  for  100  years  ? 

Ans.  $  222.25. 

15.  What  is  the  amount  of  $  76.47  for  7  years  ? 

Ans.  $  108.58,7 

16.  What  is  the  amount  of  $  716.57  for  4  years  ? 

Ans.  $  888.54,6. 

17.  What  is  the  amount  of  $  178.56,5  for  30  years  ? 

Ans.  $499.98,2. 

18.  What  is  the  interest  of  $  97.06  for  9  years  ? 

Ans.  $52.41,2. 

19.  What  is  the  interest  of  $  0.75  for  75  years  ? 

Ans.  $3.37,5. 

20.  What  is  the  interest  of  $  750  for  12  years  ? 

Ans.  $  540. 
CASE  II. 
To  find  the  interest  for  months  at  6  per  cent. 

RULE.  —  Multiply  the  principal  by  half  the  number  of  months,  ex- 
pressed decimally  as  a  per  cent.  ;  that  is,  for  12  months  multiply  by  .06  ; 
for  8  months  multiply  by  .04  ;  for  7  months,  .035  ;  for  1  month,  .005 ; 
and  point  for  decimals  as  in  the  last  rule. 

NOTE  1.  —  It  is  obvious,  that,  if  the  rate  per  cent,  were  12,  it  would  be 
1  per  cent  a  month.  If,  therefore,  it  be  6  per  cent,  it  will  be  a  half  per 
cent,  a  month,  that  is,  half  the  months  will  be  the  per  cent. 

NOTE  2.  —  If  any  other  per  cent,  is  wanted,  proceed  as  above,  and  then 
multiply  by  the  given  rate  per  cent,  and  divide  by  6,  and  the  quotient  is 
the  interest. 

1.  What  is  the  interest  of  $  368  for  8  months  ? 

$368 

.04  =  half  the  months. 
$  14.72  =  Answer. 

2.  What  is  the  interest  of  $  637  for  10  months  ? 

Ans.  $  31.85. 

3.  What  is  the  interest  of  $1671.32  for  14  months  ? 

Ans.  $116.99. 

4.  What  is  the  interest  of  $  891.24  for  9  months  ? 

Ans.  $40.10. 

5.  What  is  the  interest  of  $  819.75  for  11  months  ? 

Ans.  $  45.08,6. 

6.  What  is  the  interest  of  $  3671.25  for  13  months  ? 

Ans.  $238.63. 

7.  What  is  the  interest  of  $  61.18  for  15  months  ? 

Ans.  $  4.58. 

8.  What  is  the  interest  of  $  3181.29  for  18  months  ? 

Ans.  $  286.31. 


SECT,  xxxvn.]  SIMPLE  INTEREST.  167 

9*  What  is  the  interest  of  $  11.39  for  19  months  ? 

Ans.  $  1.08. 

10.  What  is  the  interest  of  $  9.98  for  23  months  ? 

Ans.  $  1.14. 

11.  What  is  the  interest  of  $87.19  for  27  months  ? 

Ans.  $11.77. 

12.  What  is  the  interest  of  $  32.18  for  36  months  ? 

Ans.  8  5.79. 

13.  What  is  the  interest  of  $  167.18  for  50  months  ? 

Ans.  $41.79. 

14.  What  is  the  interest  of  $  386.19  for  100  months  ? 

Ans.  $  193.09. 

CASE  III. 

To  find  the  interest  of  any  sum  for  months  and  days,  at  6 
per  cent. 

RULE.*  —  Find  the  interest  for  the  number  of  months,  as  under  Case 

II.  Then  to  find  it  for  the  number  of  days,multiply  the  principal  by  one 
sixth  the  number  of  days,  and,  if  the  principal  be  dollars,  cut  off  three 
figures  from  the  right  hand,  and  those  at  die  left  ivill  be  the  interest  in 
dollars,  and  those  at  the  right  will  be  cents  and  mills.     But  if  the  prin- 
cipal be  dollars  and  cents,  five  figures  must  be  cut  off  from  the  right 
hand,  and  those  at  the  left  will  be  the  interest,  <^c.,  as  before. 

*  The  reason  for  this  rule,  so  far  as  it  relates  to  the  interest  for  any  num- 
ber of  months,  was  explained  above.  But  the  reason  for  the  operation  in 
the  case  of  days  is  not  so  obvious.  It  will  be  seen,  however,  that  it  is 
nothing  but  an  abridgment  of  the  general  rule  for  calculating  interest,  as 
given  on  page  164,  which  will  appear  from  what  follows. 

To  find  the  interest  for  any  number  of  years,  we  multiply  the  principal 
by  the  annual  per  cent.,  and  the  product  thus  obtained  by  the  number  of 
years,  and  cut  off  two  figures,  &c.  In  like  manner,  it  is  evident  that  to 
find  the  interest  for  any  number  of  days,  we  have  only  to  multiply  by 
the  daily  per  cent.,  and  the  product  thus  obtained  by  the  number  of  days, 
and  cut  off  as  in  the  case  of  years.  But  6  per  cent,  per  annum  is  55  per 
cent  per  diem,  allowing  360  days  to  a  year.  Now,  to  multiply  the  princi- 
pal by  c  of  the  days,  and  cut  off  one  figure  from  the  product  at  the  right,  is 
the  same  as  to  multiply  by  55  (the  daily  per  cent.),  and  the  product  thus 
obtained  by  the  whole  number  of  days.  Where  the  principal  is  dollars 
only,  the  rule  directs  to  cut  off  three  figures.  Two  of  them  are  cut  off  ac- 
cording to  the  general  rule,  on  page  164,  and  the  other,  because  in  the  op- 
eration we  have  multiplied  by  |  the  number  of  days,  instead  of  $$  of  them, 
which  would  have  been  the  proper  multiplier. 

The  above  may  be  illustrated  conciselv  by  the  following  operation.  Let 
it  be  required  to  find  the  interest  of  75  dollars  for  180  days.  By  the  gen- 
eral rule  we  have  75  x  go  -*- 100  X  180  =  $  2.25.  By  the  rule  under  Case 

III.  we  have  75  X  s  of  180  =  $2.25. 


168  SIMPLE   INTEREST.  [SECT,  xixrn 

NOTE  1.  —  If  one  half  the  number  of  months  be  expressed  by  a  single 
figure,  we  have  only  to  annex  to  this  figure  |  the  number  of  days,  and 
multiply  the  principal  by  the  number  thus  found,  cutting  off  as  above, 
and  we  obtain,  by  a  single  operation,  the  interest  for  the  months  and  days. 

NOTE  2.  —  If  any  other  per  cent,  than  6  is  given,  we  may  proceed  as 
above,  and  then  multiply  by  the  given  rate  and  divide  by  6,  and  the  result 
will  be  the  interest  sought. 

1.  What  is  the  interest  of  $  68.25  for  8  months  and  24  days? 

Ans.  $3.00,3. 

OPERATION. 

$68.25 

.044        The  first  4  in  the  multiplier  is  half  of  the  8 

27300     months  ;  the  second  4  is  one  sixth  of  the  24 

27300 


2.  What  is  the  interest  of  $  637.28  for  17  months  and   19 
days,  at  8  per  cent.  ?  Ans.  $74.91,5. 

3.  What  is  the  interest  of  $396.15  for  13  months  and  9 
days  ?  Ans.  $  26.34,3. 

4.  What  is  the  interest  of  $  16.75  for  7  months  and  17  days, 
at  7  per  cent.  ?  Ans.  $  0.73,9. 

5.  What  is  the  interest  of  $976.18  for  29  months  and  23 
days,  at  9  per  cent.  ?  Ans.  $  217.93,2. 

6.  What  is  the  interest  of  $  36.18  for  3  months  and  7  days  ? 

Ans.  $0.58,4. 

7.  What  is  the  interest  of  $  51.17  for  9  months  and  29  days, 
at  4  per  cent.  ?  Ans.  $  1.69,9. 

8.  What  is  the  interest  of  $365.19  for  33  months  and  4 
days,  at  2  per  cent.  ?  Ans.  $  20.16,6. 

9.  What  is  the  interest  of  $  125.75  for  5  months  and  4 
days  ?  Ans.  $  3.22,7. 

10.  What  is  the  interest  of  $  35.49  for  1  month  and  2  days, 
at  7£  per  cent.  ?  Ans.  $  0.23,6. 

11.  What  is  the  interest  of  $  112.50  for  3  months  and  1  day, 
at  9£  per  cent.  ?  Ans.  $  2.70,1. 

12.  What  is  the  interest  of  $97.15  for  35  months  and  27 
days  ?  Ans.  $  17.43,8. 

13.  What  is  the  interest  of  $47.15  for  1  month  and  19  days, 
at  13£  per  cent.  ?  Ans.  $  0.86,6. 

14."  What  is  the  interest  of  $  678.75  for  87  -months  and  20 
days?  Ans.  $  297.51,8. 

15.  What  is  the  interest  of  $  86  for  99  months  and  29  days, 
at  25  per  cent.  ?  Ans.  $  179.10,6. 


SBCT.  xxxrii.]  SIMPLE  INTEREST.  169 

16.  What  is  the  interest  of  $33.35,8  for  15  months  and  17 
days  ?  Ans.  $  2.59,6. 

17.  What  is  the  interest  of  $  144  for  5  days  ? 

Ans.  $0.12,0. 

CASE  IV. 

When  the  interest  is  required  on  any  sum,  from  a  certain 
day  of  the  month  in  a  year,  to  a  particular  day  of  a  month  in 
the  same,  or  in  another  year. 

RULE.  —  Find  the  time,  by  placing  the  latest  date  in  an  upper  line, 
and  the  earliest  date  under  it.  Let  the  year  be  placed  first ;  the  number 
of  months  that  have  elapsed  since  the  year  commenced  at  its  right  hand, 
and  the  day  of  the  month  next ;  tJien  subtract  the  earlier  from  the  latest 
date,  and  the  remainder  is  the  time  for  which  the  interest  is  required. 
Then  proceed  as  in  the  last  rule.  Or,  the  months  may  be  reckoned  by 
their  ordinal  number,  as  in  Operation  Second. 

NOTE.  —  Many  practical  men  prefer  reckoning  interest  by  the  second 
method. 

EXAMPLES. 

1.  What  is  the  interest  of  $84.97,  from  Sept.  25,  1833,  to 

March  8,  1835  ?  Ans.  $  7.40,6. 

Operation  First.             Operation  Second.  First  Method. 

Yrs.        m.     d.                         Yre.        m.     d.  &  04.  07 

1835  2  8       1835  3  8  7is7i 

1833  8  25       1833  9  25  __'^il 

\ — iTT^       i — K~T^  59479 

1  5  13          67976 

1416 


$7.40,655 

Second  Method.  It  is  evident  that  4  months'  interest 

$  84.97  is  -^  of  a  year's  interest ;   and  for  the 

•06  same  reason,  1  month's  interest  is  £  of 

1  year,         =  5.0982  4  months'  interest ;  and  as  10  days  is  * 

4  months,  ^  ==.  1.6994  of  a  month,  the  interest  for  that  time  is 

1  month,    £  —    .4248  £  "of  a  month's  interest ;  and  if  the  in- 

10  days,    \  =    .1416  terest  of  10  days  be  divided  by  5,  the 

2  days,    £  =       283  quotient  will  be  2  days'  interest,  and 

1  day,      £  =       141  the  half  of  this  will  be  1  day's  interest. 

$7^40,64,  interest  as  before. 

2.  What  is  the  interest  of  $786.75,  from  Dec.  9,  1831,  to 

May  11,  1833  ?  Ans.  $  67.13,6. 
15 


170  SIMPLE   INTEREST.  [SKCT.  xxxvn. 

3.  What  is  the  interest  of  $  98.25,  from  July  4,  1826,  to 
Oct.  19,  1829  ?  Ans.  $  19.40,4. 

4.  What  is  the  interest  of  $76.89,5,  from  Jan.  11,  1822,  to 
July  27,  1833  ?  Ans.  $  53.26,2. 

5.  What  is  the  interest  of  $22.76,3,  from  Feb.  19,  1806,  to 
July  18,  1830  ?  Ans.  $33.34,4. 

6.  What  is  the  interest  of  $  76.35,  from  August  17,  1830, 
to  May  5,  1832  ?  Ans.  $7.86,4. 

7.  What  is  the  interest  of  $97.86,  from  May  17,  1821,  to 
Dec.  19,  1828  ?  Ans.  $  44.55,8. 

8.  What  is  the  interest  of  $  1728.75,  from  Nov.  19,1823,  to 
June  18,  1826  ?  Ans.  $267.66,8. 

9.  What  is  the  interest  of  $  99.99,9,  from  Jan.  1,  1800,  to 
Feb.  29,  1832  ?  Ans.  $  192.96,4. 

10.  What  is  the  interest  of  $  16.76,  from  Dec.  17,  1811,  to 
June  17,  1822  ?  Ans.  $10.55,8. 

11.  What  is  the  interest  of  $35.61,  from  Nov.  11,  1831,  to 
Dec.  15,  1833  ?  Ans.  $4.47,4. 

12.  What  is  the  interest  of  $  786.97,  from  Oct.  19,  1827,  to 
August  17,  1831,  at  7£  per  cent.  ?  Ans.  $  225.92,5. 

13.  What  is  the  interest  of  $96.84,  from  Nov.  27,  1829,  to 
July  3,  1832,  at  7£  per  cent  ?  Ans.  $  18.88,3. 

14.  What  is  the  interest  of  $  11.10,5,  from  April  17, 1832,  to 
Dec.  7,  1832,  at  7  per  cent.  ?  Ans.  $  0.49,6. 

15.  What  is  the  interest  of  $  117.21,  from  June  19,  1806,  to 
June  17,  1819,  at  8£  per  cent.  ?  Ans.  $  129.46,1. 

16.  What  is  the  interest  of  $  17869.75,  from  Feb.  7,  1830,  to 
Jan.  11,  1832,  at  5  per  cent.  ?  Ans.  $  1722.44,5. 

17.  What  is  the  interest  of  $71.09,1,  from  July  29,  1823,  to 
June  19,  1827,  at  12  per  cent.  ?  Ans.  $  33.17,5. 

18.  What  is  the  interest  of  $  83.47,  from  Nov.  8,  1830,  to  Ju- 
ly 11,  1833,  at  8f  per  cent.  ?  Ans.  $  19.53,7. 

19.  What  is  the  interest  of  $79.25,  from  Dec.  8,  1831,  to 
July  17,  1833?  Ans.  $7.64,7. 

20.  What  is  the  interest  of  $  175.07,  from  Jan.  7,  1825,  to 
Oct.  12,  1829  ?  Ans.  $  50.04. 

21.  What  is  the  interest  of  $  12.75,  from  June  16,  1831,  to 
August  20,  1833  ?  Ans.  $  1.66,6. 

22.  What  is  the  interest  of  $  197.28,5,  from  Dec.  6,  1932,  to 
Jan.  11,  1834  ?  Ans.  $  12.98,7. 

23.  What  is  the  interest  of  $  12.69,  from  Jan.  2,  1833,  to 
August  30,  1834,  at  7  per  cent.  ?  Ans.  $  1.47,5. 

24.  What  is  the  interest  of  $79.15,  from  Feb.  11,  1831,  to 
June  10,  1833,  at  7±  per  cent.  ?  Ans.  $  13.37,3. 


SECT,  xxxvii.]  SIMPLE  INTEREST.  171 

25.  What  is  the  amount  of  $83.33,  from  March  11, 1831,  to 
Jan.  1,  1833,  at  7£  per  cent.  ?  Ans.  $94.61,4. 

26.  What  is  the  amount  of  $100.25,  from  March  2,  1831,  to 
June  1,  1831,  at  4  per  cent.  ?  Ans.  8101.24,1. 

27.  What  is  the  amount  of  8  369.29,  from  April  30,  1830,  to 
July  31,  1832,  at  9  per  cent.  ?  Ans.  8  444.16,3. 

28.  What  is  the  interest  of  8769.87,  from  Jan.  1,  1830,  to 
June  17,  1835,  at  9£  per  cent.  ?  Ans.  8  399.41,2. 

29.  What  is  the  interest  of  869.75,  from  Jan.  11,  1833,  to 
June  29,  1833,  at  17£  per  cent.  ?  Ans.  8  5.69,6. 

30.  What  is  the  interest  of  8  368.18,  from  April  2,  1816,  to 
June  19,  1835,  at  2  per  cent.  ?  Ans.  8141.48,3. 

31.  What  is  the  interest  of  816.16,  from  March  3,  1831,  to 
Dec.  6,  1833,  at  1  per  cent.  ?  Ans.  8  0.44,5. 

32.  What  is  the  interest  of  81728.19,  from  May  7,  1824,  to 
July  17,  1830,  at  £  per  cent.  ?  Ans.  $  26.76,2. 

33.  What  is  the  interest  of  8397.16,  from  Dec.  29,  1831,  to 
June  30,  1833,  at  5£  per  cent.  ?  Ans.  8  32.82,6. 

34.  What  is  the  amount  of  $1760.07,  from  Feb.  17,  1831,  to 
Dec.  19,  1832,  at  $  per  cent.  ?  Ans.  81776.25,2. 

35.  G.  K.  M.,  of  Bradford,  has  sent  shoes  at  several  times 
to  Spofford  &  Tileston,  New  York,  as  follows :  — 

January    16,  1834,  were  sent  shoes  to  the  value  of  8  865.00 
Feb.  17,  1834,  "  "  "         386.27 

March  29,  1834,  "  "  "         769.25 

May  25,  1834,  "  "  "         183.75 

June  19,  1834,  "  "  "         396.81 

The  above  were  sold  on  six  months'  credit.  G.  K.  M.  has  re- 
ceived of  Spofford  &  Tileston  as  follows  :  — Sept.  1,  1834, 
81000;  Oct.  19,  1834,  8375.25;  Nov.  15,  1834,  $681.29; 
Dec.  8,  1834,  8100  ;  March  12,  1835,  8275.28.  Required  the 
balance  at  the  time  of  settlement,  Sept.  9,  1835. 

Ans.  Spofford  &  Tileston  owe  to  G.  K.  M.  $195.51+. 

CASE  V. 

To  find  the  interest  for  any  sum  for  days. 
RULE.*  —  If  the  rate  per  cent. be. 06, multiply  the  principal  by  the 
number  of  days,  and  divide  by  6083J,  and  the  quotient  is  the  interest ; 
but  if  the  rate  per  cent,  be  .05,  divide  by  7300. 

*  This  rule  is  but  an  abridgment  of  the  obvious  method  for  obtaining 
the  interest  for  any  number  of  days,  which  is,  first  to  find  it  for  one  year, 
then  divide  by  365,  which  gives  it  fbr  one  day,  and  then  multiply  the 


172  SIMPLE  INTEREST.  [SECT,  XMTII. 

EXAMPLES. 

1.  What  is  the  interest  of  $1835  for  35  days  ? 

Ans.  $10.55,7. 

2.  What  is  the  interest  of  $165.37  for  165  days? 

Ans.  $  4.48,5. 

3.  What  is  the  interest  of  $16.87  for  79  days,  at  5. per 
cent.?  Ans.  $0.18,2. 

4.  What  is  the  interest  of  $167  for  87  days,  at   5  per 
cent.?  Ans.  $1.99. 

5.  What  is  the  interest  of  $  761.81  for  165  days  ? 

Ans.  $  20.66,2. 

6.  What  is  the  interest  of  $76.18,5  for  315  days,  at  5  per 
cent.  ?  Ans.  $  3.28,7. 

7.  What  is  the  interest  of  $178.69,7  for  271  days  ? 

Ans.  $  7.96. 

8.  What  is  the  interest  of  $1728.79  for  318  days,  at  5  per 
cent  ?  Ans.  $  75.30,8. 

9.  What  is   the   interest  of  $73   for  73  days,  at   5   per 
cent.  ?  Ans,  $  0.73. 

10.  What  is  the  interest  of  $  96.10  for  54  days  ? 

Ans.  $  0.85,3. 

11.  What  is  the  interest  of  $144.50  for  144  days,  at  5  per 
cent.  ?  Ans.  6  2.85. 

12.  What  is  the  amount  of  $1728  for  200  days  ? 

Ans.  $1784.81. 

result  by  the  number  of  days.  The  full  operation  of  finding  the  interest 
of  any  sum,  say  $1835  for  3o  days,  will  be  as  follows: 

18353(^-06  X  35  =  10.55,7+. 

Now  it  is  evident,  that,  instead  of  multiplying  1835  (the  numerator)  by 
.06,  we  may  divide  365  (the  denominator)  by  it,  without  affecting  the  re- 
sult. But  365 -7- .06  =  6083-+-,  so  that  we  have  only  to  multiply  the 
principal  by  the  number  of  days,  and  divide  the  product  by  6083J,  when 
the  rate  is  o  per  cent.,  and  we  have  the  interest  required.  For  the  same 
reason,  we  divide  by  7300  when  the  rate  is  5  per  cent.  In  the  same 
way  a  divisor  may  be  found,  answering  to  any  given  rate. 

All  the  foregoing  rules  are  on  the  principle,  that  there  are  only  360 
days  in  a  year ;  yet  they  are  adopted  by  all  mercantile  men,  and  also  by 
banks.  But  if  a  note  be  written  for  144  days,  the  holder  of  it  can  obtain 
interest  for  only  J^f  of  a  year  by  the  law  of  Massachusetts. 

If  a  note  be  written  for  months,  calendar  months  are  understood, 
whether  the  months  have  28  or  31  days. 

If  a  note  be  dated  April  21st,  for  one  month,  it  will  be  due  May  21st; 
but  if  it  be  dated  Jan.  28th,  29th,  30th,  or  31st,  it  being  for  one  month, 
it  will  be  due  Feb.  28th,  it  being  the  last  day  of  the  month,  unless  it 
be  leap  year. 


SECT,  xxxviii.]  SIMPLE  INTEREST. 

SECTION  XXXVIII. 

PARTIAL  PAYMENTS. 

WHEN  notes  are  paid  within  one  year  from  the  time  they  be- 
come due,  it  has  been  the  usual  custom  to  find  the  amount  of 
the  principal  from  the  time  it  became  due  until  the  time  of 
payment ;  and  then  to  find  the  amount  of  each  indorsement 
from  the  time  it  was  paid  until  settlement,  and  to  subtract  their 
sum  from  the  amount  of  the  principal. 

EXAMPLES. 
(1.)     $1728.00.  Baltimore,  January  1,  1833. 

For  value  received,  I  promise  Riggs,  Peabody,  &  Co.  to  pay 
them,  or  order,  on  demand,  one  thousand  seven  hundred  and 
twenty-eight  dollars,  with  interest.  John  Paywell,  Jr. 

On  this  note  are  the  following  indorsements.  March  1,  1833,  received 
three  hundred  dollars.  May  16,  1833,  received  one  hundred  and  fifty 
dollars.  Sept.  1, 1833,  received  two  hundred  and  seventy  dollars.  De- 
cember 11,  1833,  received  one  hundred  and  thirty-five  dollars. 

What  was  due  at  the  time  of  payment,  which  was  Decem- 


ber  16,  1833  ? 

OPERATION. 

Principal,  ... 

Interest  for  11  months  and  15  days, 


First  payment,         -                          -  $300.00 

Interest  for  9  months  and  15  days,  •  14.25 

Second  payment,     ...  150.00 

Interest  for  7  months,     -             -  5.25 

Third  payment,        -                           -  270.00 

Interest  for  3  months  and  15  days,  4.72 

Fourth  payment,      ...  135.00 

Interest  for  5  days,           -             -  .11 


Ans.  $  948.03. 

$1728.00 
99.36 
$1827.36 


$  879.33 


Balance  remaining  due,  Dec.  16,  1833,     $  948.03 

(2.)     $700.00.  Concord,  Feb.  4,  1834. 

For  value  received,  we  jointly  and  severally  promise  James 
Thomas  to  pay  him,  or  order,  on  demand,  seven  hundred  dol- 
lars, with  interest.  Sampson  Phillips. 

Attest,  Henry  Dix.  Richard  Fletcher. 

15* 


174  SIMPLE  INTEREST.  [SECT,  xxxvni. 

On  this  note  are  the  following  payments.  March  18,  1834,  received 
one  hundred  and  sixty  dollars.  June  24,  1834,  received  two  hundred 
dollars.  September  11,  1834,  received  one  hundred  and  twenty  dollars. 
October  5,  Iti34,  received  sixty  dollars. 

What  was  due  on  this  note  Nov.  28,  1834  ?    Ans.  $  180.43. 

(3.)     $  600.00.  Portland,  May  16,  1834. 

For  value  received,  we,  Luke  A.  Homer,  as  principal,  and 
Daniel  D.  Snow  and  Ichabod  Frost,  as  sureties,  promise  John 
Webster  to  pay  him,  or  order,  in  one  year,  six  hundred  dollars, 
with  interest  after  three  months.  Luke  A.  Homer. 

Daniel  D.  Snow. 

Attest,  M.  Peters.  Ichabod  Frost. 

On  this  note  are  the  following  indorsements.  Sept.  18,  1834,  received 
one  hundred  and  'thirty-six  dollars.  December  5,  1834,  received  one 
hundred  and  ninety-seven  dollars.  February  11,  1835,  received  two 
hundred  dollars.  April  19,  1835,  received  forty  dollars. 

What  was  due  August  1,  1835  ?  Ans.  $  40.31,2. 

In  the  United  States  courts,  and  in  most  of  the  courts  of  the 
several  States,  the  following  rule  is  adopted  for  estimating  in- 
terest on  notes  and  bonds,  when  partial  payments  have  been 
made. 

RULE.  —  Compute  the  interest  on  the  principal  sum,  from  the  time 
when  the  interest  commenced  to  the  first  time  when  a  payment  was 
made,  which  exceeds,  either  alone  or  in  conjunction  with  the  preceding 
payments,  if  any,  the  interest  at  that  time  due ;  add  that  interest  to  the 
principal,  and  from  the  sum  subtract  the  payment  made  at  that  time, 
together  with  the  preceding  payments,  if  any,  and  the  remainder  forms 
a  new  principal ;  on  whicJi  compute  and  subtract  the  interest,  as  upon 
the  first  principal,  and  proceed  in  the  same  manner  to  the  time  of 
judgment. 

The  following  example  will  illustrate  the  above  rule. 
(4.)     $165.18.  Boston,  June  17,  1827. 

For  value  received,  I  promise  James  E.  Snow  to  pay  him, 
or  order,  on  demand,  one  hundred  and  sixty-five  dollars  and 
eighteen  cents,  with  interest.  James  Y.  Frye. 

Attest,  John  True. 

On  this  note  are  the  following  indorsements.  December  7,  1827, 
received  eighteen  dollars  and  thirteen  cents  of  the  within  note.  October 
19,  1828,  received  twenty-eight  dollars  and  sixteen  cents.  September  25, 
1829,  received  thirty-six  dollars  and  twelve  cents.  July  10,  1830,  re- 
ceived three  dollars  and  eighteen  cents.  June  6, 1831,  received  thirty- 


SECT.  XXXVIII.] 


SIMPLE  INTEREST. 


175 


six  dollars  and  twenty-eight  cents.  December  28, 1832,  received  thirty- 
one  dollars  and  seventeen  cents.  May  5,  1833,  received  three  dollars 
and  eighteen  cents.  September  1,  1833,  received  twenty-five  dollars  and 
eighteen  cents.  October  18,  1834,  received  ten  dollars. 

How  much  remains  due  September  27,  1835  ? 

Ans.  $15.41,7. 


METHOD  OF  OPERATION. 


Principal,  carrying  interest  from  June  17,  1827, 

Interest  from  June  17,  1827,  to  Dec.  7,  1827,  5mo.  20  days, 

Amount, 


First  payment,  Dec.  7,  1827, 

Balance  for  new  principal,        - 

Interest  from  Dec.  7,  1827,  to  Oct.  19,  1828,  lOmo.  12da., 

Amount, 
Second  payment,  Oct.  19,  1828,     - 

Balance  for  new  principal,       - 

Interest  from  Oct.  19,  1828,  to  Sept.  25,  1829,  llmo.  6da., 

Amount, 
Third  payment,  Sept.  25,  1829,     - 

Balance  for  new  principal,       - 

Interest  from  Sept.  25,  1829,  to  June  6, 1831,  20mo.  llda., 

Amount, 

Fourth  pay't,  July  10,  1830,  a  sum  less  than  int'st,    3.18 
Fifth  pay't,  June  6, 1831,  a  sum  greater  than  irit'st,  36.28 

Balance  for  new  principal,        - 

Interest  from  June  6, 1831,  to  Dec.  28, 1832,  I8mo.  22da., 

Amount, 
Sixth  payment,  Dec.  28,  1832,      ... 

Balance  for  new  principal, 

Interest  from  Dec.  28,  1832,  to  May  5,  1833,  4mo.  7da., 

Amount, 
Seventh  payment,  May  5,  1833,    - 

Balance  for  new  principal,       - 

Interest  from  May  5,  1833,  to  Sept.  1,  1833,  3mo.  26da., 

Amount, 
Eighth  payment,  Sept.  1,  1833,     - 

Balance  for  new  principal,        -  - 

Interest  from  Sept.  1,  1833,  to  Oct.  18,  1834, 13mo.  17da., 

Amount, 


$165.18,0 
4.68,0 

169.86,0 
18.13,0 

151.73,0 

7.88,9 

159.61,9 
28.16,0 

131.45,9 
7.36,1 

138.82,0 
36.12,0 

102.70,0 
10.45,8 

113.15,8 


39.46,0 

73.69,8 
6.90,3 

80.60,1 
31.17,0 

49.43,1 
1.04,6 

50.47,7 
3.18,0 

47.29,7 
91,4 

48.21,1 
25.18,0 

23.03,1 
1.56,2 

24.59,3 


176  SIMPLE  INTEREST.  [SECT,  xxxvm. 

Amount  brought  forward,     $  24.59,3 
Ninth  payment,    - 

Balance  for  new  principal,       -                                                    14.59,3 
Interest  from  Oct.  18,  1834,  to  Sept.  27, 1835,  llmo.  9da., .82,4 

Balance  due  at  the  time  of  payment,  -  -  $  15.41,7 

(5.)  $  769.87.  Salem,  June  17,  1829. 

For  value  received,  I  promise  L.  Swan  to  pay  him,  or  order, 
on  demand,  seven  hundred  and  sixty-nine  dollars  and  eighty- 
seven  cents,  with  interest.  Samuel  Q.  Peters.  , 

Attest,  Moses  Haynes. 

On  this  note  are  the  following  payments.  March  1, 1830,  received 
seventy-five  dollars  and  fifty  cents.  June  11,  1831,  received  one  hundred 
and  sixty-five  dollars.  September  15,  1831,  received  one  hundred  and 
sixty-one  dollars.  Jan.  21, 1832,  received  forty-seven  dollars  and  twenty- 
five  cents.  March  5,  1833,  received  twelve  dollars  and  seventeen  cents. 
December  6,  1833,  received  ninety-eight  dollars.  July  7, 1834,  received 
one  hundred  and  sixty -nine  dollars. 

What  remains  due  Sept.  25,  1835  ?          Ans.  $  226.29,7. 

(6.)     $  300.00.  Haverhill,  April  30,  1831. 

For  value  received,  I  promise  Kimball  &  Hammond  to  pay 
them,  or  order,  on  demand,  three  hundred  dollars,  with  in- 
terest. Simpson  W.  Leavet. 

Attest,  James  Quintire. 

The  following  partial  payments  were  made  on  this  note.  June  27, 
1832,  received  one  hundred  and  fifty  dollars.  December  9,  1832,  re- 
ceived one  hundred  and  fifty  dollars. 

•What  was  due,  Oct.  9,  1833  ?  Ans.  $  26.73,5. 

(7.)     $  54.18.  New  York,  Feb.  11,  1832. 

For  value  received,  I  promise  John  Trow  to  pay  him,  or 
order,  on  demand,  fifty-four  dollars  and  eighteen  cents,  with 
interest.  Luke  M.  Sampson. 

On  this  note  are  the  following  payments.  July  11,  1833,  received 
twelve  dollars  and  twenty-five  cents.  August  15,  1834,  received  two 
dollars  and  ten  cents.  July  9,  1835,  received  three  dollars  and  twelve 
cents.  August  21, 1835,  received  thirty-seven  dollars  and  eighteen  cents. 

What  was  due  Dec.  17,  1835.  Ans.  $10.22,2. 

(8.)     $1728.00.  Boston,  Jan.  7,  1831. 

For  value  received,  we  jointly  and  severally  promise  Jones, 
Oliver,  &  Co.  to  pay  them,  or  order,  on  demand,  one  thousand 
seven  hundred  and  twenty-eight  dollars,  with  interest. 

John  Bountiful. 

Attest,  Timothy  True.  James  Trusty.     . 


SECT,  xxxviii.]  SIMPLE  INTEREST.  177 

On  this  note  are  the  following  payments.  February  9, 1832,  received 
seven  hundred  and  sixty  dollars  and  twenty-eight  cents  and  five  mills. 
March  5,  1833,  received  sixty-eight  dollars  and  fifty  cents.  December 
28, 1833,  received  eight  hundred  and  seventy-six  dollars  and  twenty-eight 
cents.  July  17,  1834,  received  sixty  dollars. 

What  was  due  at  the  time  of  payment,  which  was  Oct.  1, 
1834  ?  Ans.  $  209.22,9. 

(9.)     $  500.00.  Philadelphia,  May  7,  1829. 

For  value  received,  I  promise  John  Jordan  to  pay  him,  or 
order,  on  demand,  five  hundred  dollars,  with  interest. 

Thomas  C.  True. 

The  following  partial  payments  were  indorsed  on  this  note.  June 
29,  1830,  received  one  hundred  dollars.  December  5,  1831,  received  one 
hundred  dollars.  March  12,  1832,  received  five  dollars.  July  4,  1833, 
received  ninety-five  dollars.  December  1,  1834,  received  two  hundred 
dollars. 

What  was  due  Jan.  1,  1836  ?  Ans.  $141.50,4. 

(10.)     $  89.75.  Newburyport,  March  19,  1831. 

For  value  received,  we  jointly  and  severally  promise  John 
Frost  to  pay  him,  or  order,  on  demand,  eighty-nine  dollars  and 
seventy- five  cents,  with  interest.  Henry  Augustus. 

Attest,  James  Snow.  Marcus  T.  Cicero. 

On  this  note  are  the  following  indorsements.  December  6,  1831,  re- 
ceived twelve  dollars  and  twelve  cents.  February  17,  1832,  received 
twelve  dollars  and  twelve  cents.  March  19,  1833,  received  three  dollars 
and  sixteen  cents.  December  28,  1834,  received  two  dollars  and  eighteen 
cents.  January  1,  1835,  received  twenty-five  dollars  and  twenty-five 
cents.  March  11,  1835,  received  thirty-one  dollars  arid  eighteen  cents. 
July  17,  1835,  received  five  dollars  and  eighteen  cents.  September  1, 
1835,  received  six  dollars  and  twenty-nine  cents. 

What  was  due  Dec.  29,  1835  ?  Ans.  $10.57. 

(11.)     $  1000.00.  New  York,  January  1,  1840. 

For  value  received,  I  promise  to  pay  James  Johnson,  or  or- 
der, on  demand,  one  thousand  dollars,  with  interest  at  seven 
per  cent.  Samuel  T.  Fortune. 

Attest,  Job  Harris. 

On  this  note  are  the  following  indorsements.  Sept.  28, 1840,  received 
one  hundred  and  forty-four  dollars.  March  1, 1841,  received  twenty  dol- 
lars. July  17,  1841,  received  three  hundred  and  sixty  dollars.  Aug.  9, 
1841,  received  one  hundred  and  ninety  dollars.  Sept  25,  1842,  received 
one  hundred  and  seventy  dollars.  Dec.  11,  1843,  received  two  hundred 
dollars.  July  4,  1845,  received  seventy-five  dollars. 

What  was  due  June  1,  1847  ?  Ans.  $  7.61. 


178  SIMPLE  INTEREST.  [SECT,  xxxvm. 

The  following  is  the  rule  established  by  the  Supreme  Court 
of  the  State  of  Connecticut 

RULE.  —  Compute  the  interest  to  the  time  of  the  first  payment ;  if 
that  be  one  year  or  more  from  the  time  the  interest  commenced,  add  it  to 
the  principal,  and  deduct  the  payment  from  the  sum  total.  If  there  be 
after  payments  made,  compute  the  interest  on  the  balance  due  to  the 
next  payment,  and  then  deduct  the  payment  as  above;  and  in  like  man- 
ner from  one  payment  to  another,  till  all  the  payments  are  absori>ed ; 
provided  the  time  between  one  payment  and  another  be  one  year  or  more. 
But  if  any  payments  be  made  before  one  year's  interest  hath  accrued, 
then  compute  the  interest  on  the  principal  sum  due  on  the  obligation 
for  one  year,  add  it  to  the  principal,  and  compute  the  interest  on  the 
sum  paid  from  the  time  it  was  paid  up  to  the  end  of  the  year ;  add  it 
to  the  sum  paid,  and  deduct  that  sum  from  the  principal  and  interest 
added  together.* 

If  any  payments  be  made  of  a  less  sum  than  the  interest  arisen  at  the 
time  of  such  payment,  no  interest  is  to  be  computed,  but  only  on  the 
principal  sum  for  any  period. 

(12.)     $  900.00.  New  Haven,  June  1,  1828. 

For  value  received,  I  promise  J.  D.  to  pay  him,  or  order, 
nine  hundred  dollars,  on  demand,  with  interest. 

James  L.  Emerson. 

On  this  note  are  the  following  indorsements.  June  16,  1829,  received 
two  hundred  dollars  of  the  within  note.  Aug.  1,  1830,  received  one  hun- 
dred and  sixty  dollars.  Nov.  16,  1830,  received  seventy-five  dollars. 
Feb.  1,  1832,  received  two  hundred  and  twenty  dollars. 

What  was  due  August  1,  1832  ?  Ans.  $  417.82,2. 

OPERATION. 

Principal,         ......        $900.00 

Interest  from  June  1,  1828,  to  June  16,  1829,  12£  months,       56.25 

956.25 
First  payment,  .....  200.00 

756.25 
Interest  from  June  16,  1829,  to  Aug.  1,  1830,  13£  months,      51.04,6 

807.29,6 
Second  payment,         ...  .  160.00,0 

647.29,6 
Interest  for  one  year,  38.83,7 

686.13,3 

*  If  the  year  extends  beyond  the  time  of  payment,  find  the  amount  of 
the  remaining  principal  to  the  time  of  payment;  find  also  the  amount  of 
indorsement,  or  indorsements,  and  subtract  their  sum  from  the  amount 
of  the  principal. 


SECT,  XXXYIII.]  SIMPLE  INTEREST.  179 

Amount  brought  forward,         $  686.13,3 
Am't  of  3d  pay't,  from  Nov.  16, .1830,  to  Aug.  1,  1831,  8£mo.,  78.18,7 

607.94,6 
Interest  from  Aug.  1,  1831,  to  Aug.  1,  1832,  12  months,         36.47,6 

644.42,2 
Am't  of  4th  pay't,  from  Feb.  1, 1832,  to  Aug.  1, 1832,  6mo.,     226.60,0 

Balance  due  August  1,  1832,  -        $417.82,2 

Method  of  computing  annual  interest  in  the  State  of  Ver- 
mont. 

When  the  contract  is  for  the  payment  of  interest  annually, 
and  no  payments  have  been  made,  we  adopt  the  following 

RULE.  —  Find  the  interest  of  the  principal  for  each  year,  separately, 
up  to  the  time  of  payment ;  we  must  then  find  the  simple  interest  of  these 
interests,  severally ,  from  the  time  they  become  due  up  to  the  time  of  pay- 
ment, and  the  sum  of  all  the  interests  added  to  the  principal  will  be  the 
amount. 

Should  payments  have  been  made,  we  find  the  amount  of  the  principal 
and  from  this  we  subtract  the  amount  of  the  indorsement  or  indorse- 
ments to  the  end  of  the  first  year.  We  proceed  in  the  same  manner  to 
the  time  of  payment , 

But  when  the  contract  is  for  a  sum  payable  at  a  specified  time,  with, 
annual  interest,  and  payments  are  made  before  the  debt  becomes  due,  we 
find  the  interest  of  the  principal  up  to  the  time  of  the  first  payment,  and 
this  interest  we  reserve ;  we  then  subtract  the  payment  from  the  princi- 
pal, and  find  the  interest  of  the  remainder  up  to  the  time  of  the  next 
payment,  which  interest  we  add  to  the  other  interest,  and  so  continue  up 
to-  the  time  the  debt  becomes  due,  and  the  sum  of  the  interests,  added  to 
the  last  principal,  ivill  be  the  amount  due  at  that  time.  After  the  debt 
becomes  due,  the  interest  is  to  be  extinguished  annually,  if  the  payments 
are  sufficient  for  that  purpose. 

13.  J.  Jones  has  J.  Smith's  note,  dated  January  1,  1840,  for 
$  500,  with  interest,  to  be  paid  annually,  at  6  per  cent.  What 
was  due  January  1,  1844  ?  Ans.  $  630.80. 

OPERATION. 

1st  year,  $  500  x  .06  —  $  30.  $  30  X .  18  =  $  5.40 
2d  '  "  $  500  x. 06  =  $30.  $30  x  .12  =$3.60 
3d  "  $  500  x. 06  =  $30.  $30  x  .06  =  $1.80 
4th  «  $  500  X.  06  =  $30.  $1080  interest  of 

$  120  interest  of  principal.  [interest. 

Principal  $  500  00.  Th-g  .g  ^    method  when 

Interest  of  principal,       120.00.  ^  no  indorsements> 

Interest  of  interest, 

Amount,  $  630,80  Ans. 


180 


SIMPLE  INTEREST. 


[SECT.  ZXXVIH. 


(14.)     $  300.00.  Thetford,  Vt.,  January  1,  1843. 

For  value  received,  I  promise  to  pay  Hiram  Orcutt,  Esq., 
or  order,  on  demand,  three  hundred  dollars,  with  interest  an- 
nually. M.  T.  Cicero. 

Attest,  P.  M.  Virgil. 

On  this  note  are  the  following  indorsements.  Sept.  1, 1843,  received 
eighty  dollars.  July  1,  1844,  received  one  hundred  dollars.  .April  1, 
1845,  received  fifty  dollars. 

What  was  due  January  1,  1847  ? 


OPERATION. 


Principal, 

Interest  for  one  year, 


First  payment,        ... 
Interest  on  first  payment,  4  months,      - 

Amount  of  payment, 

New  principal,  ... 

Interest  on  new  principal,  one  year, 


Second  payment, 

Interest  on  second  payment,  6  months, 

Amount  of  payment,     - 

New  principal,        ... 
Interest  on  new  principal,  one  year, 


Third  payment, 

Interest  on  third  payment,  9  months, 


$300.00 
18.00 

Amount,        318.00 

80.00 
1.60 

-  sTeo 

236.40 
14.18 

Amount,        250.58 

100.00 
3.00 

103.00 

-  147.58 


Amount,, 


50.00 
2.25 

52.25 

104.18 
6.25 

-   $  1 10.43 


Amount  of  payment,     .... 

New  principal,        ..... 
Interest  on  new  principal,  one  year, 

Remains  due  January  1,  1847,        ... 

The  above  is  the  method  of  computing  annual  interest,  when 
there  are  indorsements  on  the  note,  and  it  is  not  payable  at  a 
specified  time. 

(15.)     8  600.00.  Montpelier,  Vt.,  June  1,  1844. 

For  value  received,  I  promise  to  pay  John  Smith,  or  order, 
six  hundred  dollars,  in  three  years,  with  interest  annually. 

Attest,  S.  Morse.  John  Y.  Jones. 


SECT,  xxxix.]  SIMPLE  INTEREST.  181 

On  this  note  are  the  following  indorsements.  Sept.  1, 1845,  received  two 
hundred  and  fifty  dollars.  Nov.  1,  1846,  received  two  hundred  dollars. 

What  was  due  June  1,  1847  ?  Ans.  $  242.75. 

OPERATION. 

$600  x  .105  =  63.00,  interest  from  June  1,  1844,  to  Sept.  1, 

250  [1845. 
350  X. 07  =24.50,  interest  from  Sept.  1,  1845,  to  Nov.  1, 

200  [1846. 

150  x  .035=  5.25,  interest  from  Nov.  1,  1846,  to  June  1, 

92-75  $  §2.75,  amount  of  interest.  [1847. 

$242.75  remains  due  June  1,  1847. 

The  above  is  the  method  of  computing  annual  interest,  when 
the  contract  is  for  a  specified  time. 

NOTE.  —  The  above  methods  of  computing  annual  interest  are  not  only 
practised  in  the  courts  of  Vermont,  but  a  similar  method  has  been  adopted 
in  some  of  the  courts  of  New  Hampshire  j  but  it  is  not  the  method  in 

Massachusetts. 


SECTION  XXXIX. 
MISCELLANEOUS  PROBLEMS  IN  INTEREST. 

PRINCIPAL,  interest,  and  time  given,  to  find  the  rate  per  cent. 

1.  At  what  rate  per  cent,  must  $500  be  put  on  interest  to 
gain  $  120  in  4  years  ? 

OPERATION.  BY    ANALYSIS. 

$500  The   interest  of  $1    for  the 

.0 1  given  time  at  one  per  cent,  is  4 

5  00  cents.    $  500  will  be  500  times  as 

4  much,  =  $  500  X. 04  =  $20.00. 

2000)  120.00(6  per  cent.  Ans.  ^  If.  *.201f7e  l  Per  cen*' 
120.00  ®  2   Wl11  &lve  W*  —  6  Per  cent- 

RULE.  —  Divide  the  given  interest  by  the  interest  of  the  given  sum 
at  1  per  cent,  for  the  given  time,  and  the  quotient  will  be  the  rate  per 
cent,  required. 

2.  At  what  rate  per  cent,  must  8120  be  on  interest  to  amount 
to  $133.20  in  16  months  ?  Ans.  8J-  per  cent. 

16 


182  SIMPLE  INTEREST.  [SECT,  xxxix. 

3.  At  what  rate  per  cent,  must  $  280  be  on  interest  to  amount 
to  $  411.95  in  6£  years  ?  Ans.  7£  per  cent. 

Principal,  interest,  and  rate  per  cent,  given,  to  find  the  time. 

4.  How  long  must  $500  be  on  interest  at  6  per  cent,  to 
gain  $120  ? 

OPERATION.  BY    ANALYSIS. 

$  50°  We  find  the  interest  of  8  1  .00  at 

the  given  rate  for  one  year  is  6 

30.00)  120.00(4  years,  Ans.  cents.  $  500  will,  therefore,  be  500 
120.00  times  as  much,  =  $  500  X  -06  = 

$  30.00.    Now,  if  it  take  1  year  to 
gain  $  30,  it  will  require  -^  to  gain  $  120,  =  4  years,  Ans. 


RULE.  —  Divide  the  given  interest  by  the  interest  of  the  principal  for 
1  year,  and  the  quotient  is  the  time. 

5.  How  long  must  $120  be  on  interest  at  8£  percent,  to 
amount  to  $133.20?  Ans.  16  months. 

6.  How  long  must  $  280  be  on  interest  at  7£  per  cent,  to 
amount  to  $  411.95  ?  Ans.  6£  years. 

Interest,  time,  and  rate  per  cent,  given,  to  find  the  principal. 

7.  What  principal  at  6  per  cent,  is  sufficient  in  4  years  to 
gain  $120? 

OPERATION.  BY   ANALYSIS. 

$1 

.06  The  interest  of  $1  for  the  given 

^Qg  rate  and  time  is  24  cents.     If  24 

4  cents,  then,  give  $1  for  principal, 

^4)120.00(1  500,  Ans.       $12°'00  ™*g™  1W*  times  as 
120  00  much,  —  $  500,  Ans. 

RULE.  —  Divide  the  given  interest  or  amount  by  the  interest  or 
amount  of  $  1  for  the  given  rate  and  time,  and  the  quotient  is  the 
principal. 

8.  What  principal  at  8£  per  cent,  is  sufficient  in  16  months 
to  gain  $  13.20  ?  Ans.  $  120.00. 

9.  What  principal  at  7£  per  cent,  is  sufficient  in  6J-  years  to 
amount  to  $  41  1.95  ?  Ans.  $  280.00. 


I 

SFCT.  XL.]  COMPOUND  INTEREST.  183 

SECTION  XL. 
COMPOUND  INTEREST. 

THE  law  specifies  that  the  borrower  of  money  shall  pay  a 
certain  number  of  dollars,  called  per  cent.,  for  the  use  of  one 
hundred  dollars  for  a  year.  Now,  if  this  borrower  does  not 
pay  to  the  lender  this  per  cent,  at  the  end  of  the  year,  it  is  no 
more  than  just  that  he  should  pay  interest  for  the  use  of  it,  so 
long  as  he  shall  keep  it  in  his  possession ;  and  this  is  called 
Compound  Interest. 

1.  What  is  the  compound  interest  of  $  300  for  3  years  ? 

Ans.  9  57.30,4. 

First  Method.  Second  Method. 

$  300,  principal.  5  =  §1S)300 

1.06  l=s)  15 

TOO",  interest  for  1  year.  ? 

300  5  =  ^)318 

318.00,  amount  for  1  year.  1==3  )  15.90 

1.06  3"18 


19.08,  int.  for  second  year.  !*<?*«  . 

313  1  =  1)  16.85,4 

3.37,0 
337.08,  amount  for  2  years.  3&y  3Q  4 

lt06  300* 

20.22,48,  int.  for  third  year.  ^  ^    qn  , 

OOW    AO  «JpO*»OU)^t 

337.08 

357.30,48,  amount  for  3  years.  NOTE.  —  5    per    cent,    is 

300  §5  of  the  principal,  and  1  per 

$  57.30,48,  compound  int.  for  3yrs.  5  ?er 

RULE.  — Find  the  interest  of  the  given  sum  for  one  year,  and  add 
it  to  the  principal ;  then  find  the  interest  of  this  amount  for  the  next 
year ;  and  so  continue  until  the  time  of  settlement.  Subtract  the  prin- 
cipal from  the  last  amount,  and  the  remainder  is  the  compound  in- 
terest. 

2.  WThat  is  the  amount  of  $  500  for  3  years  ? 

Ans.  $  595.50,8. 
3    What  is  the  compound  interest  of  $  345  for  10  years  ? 

Ans.  $272.84,2. 

4.  What  is  the  compound  interest  of  $316  for  3  years  4 
months  and  18  days  ?  Ans.  $  69.01,7. 


184 


COMPOUND  INTEREST. 


[SECT.  XL. 


By  the  aid  of  the  following  Table,  calculations  are  more  easily  effected 
than  by  the  preceding  rule. 

TABLE, 

Showing  the  amount  of  1  dollar,  or  1  pound,  for  any  number  of  years  un- 
der 40,  at  3,  4,  5,  6,  and  7  per  cent.,  compound  interest. 


Yeara. 

3  per  cent. 

4  per  cent. 

5  per  cent. 

6  per  cent. 

7  per  cent. 

Years. 

1 

.030000 

1.040000 

1.050000 

1.060000 

1.070000 

1 

2 

.060900 

1.181600 

1.102500 

1.123600 

1.144900 

2 

3 

.092727 

1.124864 

1.157625 

1.191016 

1.225043 

3 

4 

.125508 

1.169858 

1.215506 

1.262476 

1.310795 

4 

5 

.159274 

1.216652 

1.276281 

1.338225 

1.402552 

5 

6 

.194052 

1.265319 

1.340095 

1.418519 

1.500730 

6 

7 

1.229873 

1.315931 

1.407100 

1.503630 

1.605781 

7 

8 

1.266770 

1.368569 

1.477455 

1.593848 

1.718186 

8 

9 

1.304773 

1.423311 

1.551328 

1.689478 

1.838459 

9 

10 

1.343916 

1.480284 

1.628894 

1.790847 

1.967151 

10 

11 

1.384233 

1.539454 

1.710339 

1.898298 

2.104852 

11 

12 

.425760 

1.601032 

1.795856 

2.012196 

2.252191 

12 

13 

.468533 

1.665073 

1.885649 

2.132928 

2.409845 

13 

14 

.512589 

1.731676 

1.979931 

2.260903 

2.578534 

14 

15 

.557967 

1.800943 

2.078928 

2.396558 

2.75%32 

15 

16 

.604706 

1.872981 

2.182874 

2.540351 

2.952164 

16 

17 

.652847 

1.947900 

2.292018 

2.692772 

3.158815 

17 

18 

.702433 

2.025816 

2.406619 

2.854339 

3.379932 

18 

19 

.753506 

2.106849 

2.526950 

3.025599 

3.616528 

19 

20 

1.806111 

2.191123 

2.653297 

3.207135 

3869685 

20 

21 

1.860294 

2.278768 

2.785962 

3.399563 

4.140563 

21 

22 

1.916103 

2.369918 

2.925260 

3.603537 

4.430403 

22 

23 

1.973586 

2.464715 

3.071523 

3.819749 

4.740530 

23 

24 

2.032794 

2.563304 

3.225099 

4.048934 

5.072367 

24 

25 

2.093777 

2.665836 

3.386354 

4.291870 

5.427434 

25 

26 

2.156591 

2.772469 

3.555672 

4.549382 

5.807352 

26 

27 

2.221289 

2.883368 

3.733456 

4.822345 

6.213868 

27 

28 

2.287927 

2.998703 

3.920129 

5.111686 

6.648838 

28 

29 

2.356565 

3.118651 

4.116135 

5.418387 

7.114257 

29 

30 

2.427262 

3.243397 

4.321942 

5.743491 

7.612255 

30 

31 

2.500080 

3.373133 

4.538039 

6.088100 

8.145112 

31 

32 

2.575082 

3.508058 

4.764941 

6.453386 

8.715270 

32 

33 

2.652335 

3.648381 

5.003188 

6.840589 

9.325339 

33 

34 

2.731905 

3.794316 

5.253347 

7.251025 

9.978113 

34 

35 

2.813862 

3.946088 

5.516015 

7.686086 

10.676581 

35 

36 

2.898278 

4.103932 

5.791810 

8.147252 

11.423942 

36 

37 

2.985226 

4.268089 

6.081406 

8.636087 

12.223618 

37 

38 

3.074783 

4.438813 

6.385477 

9.154252 

13.079271 

38 

39 

3.167026 

4.616365 

6.704751 

9.703507 

13.904820 

39 

40 

3.262037 

4.801020 

7.039988 

10.285717 

14.974457 

40 

To  perform  questions  by  the  Table,  multiply  the  amount  of  one  dollar 


SECT.  XL.]  COMPOUND  INTEREST.  185 

for  the  given  rate  and  time  found  in  the  Table  by  the  principal,  and  from 
the  product  subtract  the  principal,  and  the  remainder  is  me  compound 
interest. 

NOTE.  —  If  there  be  months  and  days,  find  the  amount  for  them  on  the 
number  taken  from  the  Table,  before  it  is  multiplied  by  the  principal. 

EXAMPLES. 

5.  What  is  the  compound  interest  of  $  360  for  5  years,  6 
months,  and  24  days  ?  Ans.  $138.14. 

OPERATION. 

1.338225,  amount  of  1  dollar  for  5  years. 
.034,  ratio  for  6  months  and  24  days. 

5352900 
4014675 


.045499650 
1.338225 

1.383724650,  amount  of  1  dollar  for  5yrs.  6mo.  24  days. 
360 


83023479000 
415117395 

498.14,0874000,  amount  of  principal  for  5yrs.  6mo.  24  days. 
360 
$  138.14,  compound  interest  of  the  principal  for  do. 

6.  What  is  the  interest  of  $  890  for  30  years  ? 

Ans.  $  4221.70,6. 

7.  What  is  the  amount  of  $  480  for  40  years  ? 

Ans.  $4937.14,4. 

8.  What  is  the  interest  of  $  300  for  10  years,  7  months, 
and  15  days  ?  Ans.  $  257.40,1. 

9.  What  is  the  amount  of  $  586  for  12  years,  1  month,  and 
29  days,  at  5  per  cent.  ?  Ans.  $1060.99,5 

10.  The  probable  number  of  blacks  at  this  time  (lb«35)  in 
the  United  States  is  2,500,000.     Now,  supposing  their  increase 
to  be  25  per  cent,  for  every  10  years,  what  will  be  their  num- 
ber in  the  year  1935  ?  Ans.  23,283,037. 

11.  Supposing  the  annual  increase  of  the  people  of  color  t< 
be  2  per  cent.,  how  many  must  be  sent  out  of  the  country 
each  year,  that  their  numbers  might  not  increase  ? 

Ans.  50,000. 

12.  What  is  the  amount  of  $  900  for  7  years,  at  5  per 
cent  ?  Ans.  $1266.39. 

16* 


186  COMPOUND  INTEREST.  [SECT.  XL. 

13.  What  is  the  interest  of  $  350  for  3  years,  3  months,  and 
24  days  ?  Ans.  $  74.77,5-J-. 

14.  What  is  the  interest  of  $  970  for  2  years,  9  months,  and 
24  days  ?  Ans.  $173.29,5. 

To  find  the  amount  of  a  note  by  compound  interest,  when 
there  have  been  partial  payments. 

RULE.  —  Find  the  amount  of  the  principal,  and  from  it  subtract  the 
amount  of  the  indorsements. 

EXAMPLES. 

15.  A,  by  his  note  dated  January  1,  1830,  promises  to  pay 
B  $  500  on  demand. 

On  this  note  are  the  following  indorsements.  July  16,  1830,  received 
two  hundred  dollars.  August  21,  1831,  received  two  hundred  dollars. 
December  1,  1832,  received  one  hundred  dollars. 

What  was  the  balance  Sept.  1,  1834  ?  Ans.  $  52.73. 

(16.)     $100.00.  Boston,  Sept.  25,  1833. 

For  value  received,  I  promise  Peter  Absalom  to  pay  him,  or 
order,  on  demand,  one  hundred  dollars,  with  interest  after  six 
months.  J.  P.  Jay. 

On  this  note  are  the  following  indorsements.  June  11, 1834,  received 
fifty  dollars.  Sept.  25, 1834,  received  fifty  dollars. 

What  was  due  August  25,  1835  ?  Ans.  $  2.24,7. 

(17.)     $1000.00.  New  York,  January  1,  1840. 

For  value  received,  I  promise  to  pay  J.  R.  Montgomery,  or 
order,  on  demand,  one  thousand  dollars,  with  interest  at  seven 
per  cent.  John  Q.  Smith. 

Attest,  J.  Page. 

On  this  note  are  the  following  indorsements.  June  10, 1840,  received 
seventy  dollars.  Sept.  25, 1841,  received  eighty  dollars.  July  4,  1842, 
received  one  hundred  dollars.  Nov.  11,  1843,  received  thirty  dollars. 
June  5, 1844,  received  fifty  dollars. 

What  remains  due  April  1,  1845,  at  7  per  cent,  compound 
interest?  Ans.  $1022.34. 

(18.)     $  1700.00.  Bradford,  July  4,  1841 

Lent  A.  Brown  seventeen  hundred  dollars.     Sept.  1,  1843, 

received  one  thousand  dollars. 

What  is  due  July  4, 1847,  at  5  per  cent,  compound  interest. 

Ans.  $1071.81,9.  - 


SECT.  XLI.]  DISCOUNT.  187 

SECTION  XLI. 
DISCOUNT. 

THE  object  of  Discount  is  to  show  us  what  allowance  should 
be  made,  when  any  sum  of  money  is  paid  before  it  becomes 
due. 

The  present  worth  of  any  sum  is  the  principal  that  must  be 
put  at  interest  to  amount  to  that  sum  in  the  given  time.  That 
is,  $  100  is  the  present  worth  of  $  106  due  one  year  hence  ; 
because  8  100  at  6  per  cent,  will  amount  to  8  106,  and  $  6  is 
the  discount. 

1.  What  is  the  present  worth  of  8  12.72,  due  one  year  hence? 
First  Method.  Second  Method. 

8 12.72  $ 

100  1.06)12.72(8 12  Ans. 
106)  1272.00(8  12  Ans.  ]^_ 

106  2.12 

212  2.12 

212 

As  $  100  will  amount  to  8  106  in  one  year  at  6  per  cent.,  it 
is  evident,  that,  if  Tg£  of  any  sum  be  taken,  it  will  be  its  pres- 
ent worth  for  one  year,  and  that  y^  will  be  the  discount. 
And,  as  8  1  is  the  present  worth  of  8  1-06  due  one  year  hence, 
it  is  evident  that  the  present  worth  of  812.72  must  be  equal  to 
the  number  of  times  8  12.72  will  contain  $1.06. 

RULE.  —  Divide  the  given  sum  by  the  amount  of  $  1  for  the  given 
rate  and  time,  and  the  quotient  will  be  the  present  ivorth.  If  the  pres- 
ent worth  be  subtracted  from  the  given  sum,  the  remainder  will  be  the 
discount. 

2.  What  is  the  present  worth  of  8  117.60,  due  one  year 
hence,  at  12  per  cent.  ?  Ans.  8  105.00. 

3.  What  is  the  discount  of  8  802.50,  at  7  per  cent.,  due  one 
year  hence  ?  Ans.  8  52.50. 

4.  What  is.  the  present  worth  of  8769.60,  due  3  years  and 
5  months  hence  ?  Ans.  $  638.67,2^¥8T- 

5.  What  is  the  present  worth  of  8  986.40,  due  7  years  9 
months  and  20  days  hence  ?  Ans.  $  671.78,2/?8T. 

6.  How  much  grain  must  be  sent  to  the  miller  that  a  bush- 
el of  meal  may  be  returned,  the  miller  taking  y^  part  for  toll  ? 

Ans.  34^  quarts. 


188  PER  CENTAGE.  [SECT.  XLII. 

7.  What  is  the  present  worth  of  $  678.75,  due  3  years  7 
months  hence,  at  7£  per  cent.  ?•  Ans.  $  534.97,5^. 

8.  I  have  given  my  note  for  $  1000,  to  be  paid  Dec.  18, 
1835.     What  is  the  note  worth  June  7,  1834  ? 

Ans.  $915.89sWT- 

9.  James  has  a  note  against  Samuel  for  $715.50,  dated 
August  17,  1834,  which  becomes  due  January  11,  1835.    What 
ready  money  will  pay  the  note  Sept.  25,  1834  ? 

Ans.  $703.07,8+. 

10.  A  has  B's  note,  which  becomes  due  Nov.  25,  1835,  for 
$  914.75.     What  is  this  note  worth  Jan.  1,  1835  ? 

Ans.  $867.88,4-f. 

11.  A  merchant  has  given  two  notes ;  the  first  for  $79.87,  to 
be  paid  Jan.  21,  1836 ;  the  second  for  $  87.75,  to  be  paid  Dec. 
17,  1836.     What  ready  money  will  discharge  both  notes  Feb. 
10,  1835  ?  Ans.  $  154.54,4-f. 

12.  It  being  now  Oct.  14,  1833,  and  A  owing  me  $  1728,  to 
be   paid  Dec.   17,  1837,  what  ought  I  now  to  receive  as  an 
equivalent  ?  Ans.  $  1 38 1.84, 7^3T. 

13.  Bought  cloth  in  Boston  at  $  5.00  per  yard.     What  must 
be  my  "  asking  price,"    in  order  that  I  may  fall  on  it  10  per 
cent,  and  still  make  10  per  cent,  on  my  purchase  ? 

Ans.  $6.1  If 

14.  James  Ober  owes  Samuel  Hall  as  follows  :  $  365.87,  to 
be  paid  Dec.   19,  1835;  $  161.15,  to  be  paid  July  16,  1836; 
$  112.50,  to  be  paid  Jane  23,  1834  ;  $96.81,  to  be  paid  April 
19,  1838.     What  should  he  receive  as  an  equivalent,  Jan.  1, 
1834  ?  Ans.  $  653.40-f-. 


SECTION  XLII. 
PER   CENTAGE. 

THIS  term  is  used  to  express  so  much  by  the  hundred.  It  is 
derived  from  two  Latin  words,  per  and  centum,  and  means  by  the 
hundred.  It  is  not  only  applied  to  money,  but  to  any  commodity. 

The  process  of  operation  is  similar  to  Interest. 

EXAMPLES. 

1.  Received  a  legacy  of  $  1728.  I  gave  10  per  cent,  of  it 
to  a  benevolent  society.  How  much  had  I  remaining  ? 

Ans.  $1555.20. 


SECT.  XLIII.J  BROKERAGE.  189 

2.  Bought  a  horse  for  $120,  and  sold  him  at  6  per  cent,  ad- 
vance.    What  did  I  gain  ?  Ans.  $  7.20. 

3.  Sent  1728  barrels  of  flour  to  Liverpool,  but  in  a  storm 
25  per  cent,  of  them  were  thrown  overboard  ;  how  many  re- 
mained ?  Ans.  1296  barrels. 

4.  A  certain  colonel,  whose  regiment  consisted  of  900  men, 
lost  8  per  cent,  of  them  in  battle,  and  50  per  cent,  of  the  re- 
mainder by  sickness.     How  many  had  he  remaining  ? 

Ans.  414  men. 

5.  A  merchant,  having  $1728  in  the  Union  Bank,  wishes  to 
withdraw  15  per  cent.  ;  how  much  will  remain  ? 

Ans.  $1468.80. 

6.  A  gentleman,  who  had  an  estate  of  $  25,000,  gave  in  his 
will,  to  his  wife,  40  per  cent,  of  his  property,  and  to  his  son 
Samuel,  30  per  cent,  of  the  remainder.     The  residue  he  divid- 
ed equally  between  his  daughters,  Marcia,  Isabella,  and  Clara, 
after  having  deducted   $  60  as  a  present   to  his  clergyman. 
What  did  each  receive  ? 

Ans.  Wife,  $10,000 ;  son,  $  4,500  ;  daughters,  $  3,480  each. 

7.  What  is  15  per  cent,  on  500  bushels  ?  Ans.  75  bushels. 

8.  What  is  20  per  cent,  on  75cwt.  ?  Ans.  15cwt. 

9.  What  is  30  per  cent,  on  150  tons  ?  Ans.  45  tons. 
10.  What  is  75  per  cent,  on  $  500  ?  Ans.  $  375. 

-11.  What  is  95  per  cent,  on  700  chaldrons  ? 

Ans.  665  chaldrons. 

12.  What  is  2  per  cent,  on  40  miles  ?  •          Ans.  8  miles. 

13.  What  is  99  per  cent,  on  $1000  ?  Ans.  $990. 

14.  What  is  33£  per  cent  on  144  barrels  ?        Ans.  48bbl. 

15.  What  is  66§  per  cent,  on  90  hogsheads  ? 

Ans.  60  hogsheads. 

16.  What  is  ±  per  cent,  on  $100  ?  Ans.  $  0.25. 

17.  What  is  £  per  cent,  on  17281b.  ?  Ans.  15121b. 


SECTION  XLIII. 
COMMISSION   AND   BROKERAGE. 

COMMISSION  AND  BROKERAGE  are  compensations  made  to  fac- 
tors, brokers,  and  other  agents,  for  their  services,  either  for  buy- 
ing or  selling  goods. 

NOTE.  —  A  factor  is  an  agent,  employed  by  merchants  residing  in  other 


190  BROKERAGE.  [SECT.  XLIH. 

places  to  buy  and  sell,  and  to  transact  business  on  their  account.     A  bro- 
ker is  employed  by  merchants  to  transact  business. 

RULE.  —  T/ie  method  of  operation  is  the  same  as  in  Interest  and  Dis- 
count. 

EXAMPLES. 

1.  My  agent  in  New  Orleans  has  purchased  cotton,  on  my 
account,  to  the  amount  of  $18,768  ;  what  is  his  commission,  at 
If  per  cent.  ?  Ans.  $  328.44. 

2.  If  a  broker  sells  goods  for  me  to  the  amount  of  $  896, 
what  is  his  commission,  at  2  per  cent.  ?  Ans.  $17.92. 

3.  My  factor  in  London  writes  that  he  has  purchased  for  me 
to  the  amount  of  395<£.  15s.  5d. ;  what  is  his  commission,  at  2£ 
per  cent.  ?  Ans.  &£.  18s.  l/^d. 

4.  A  factor  receives   $1976,  which  he  is  to  lay  out  for 
goods  ;  having  deducted  his   commission  of  4  per  cent,  how 
much  will  remain  to  be  laid  out  ?  Ans.  $1900. 

NOTE.  —  As  his  commission  is  to  be  taken  out  of  the  sum  remitted  he 
will  not  receive  4  dollars  on  every  100,  but  4  on  every  1U4  ;  that  is,  he 
will  receive  104. 

5.  What  is  a  broker's  commission  for  purchasing  goods  to 
the  amount  of  $  7658.75,  at  1£  per  cent  ?      Ans.  $114.88£. 

6.  My  factor  at  New  Orleans  advises  me  that  he  has  pur- 
chased on  my  account  37  bales  of  cotton,  at  $107.75  per  bale  ; 
what  is  his  commission,  at  f  per  cent.  ?  Ans.  $14.95^. 

7.  I  have  engaged  a  broker  to  purchase  for  me  12  shares  in 
the  Boston  and  Mairie  Railroad,  at  $  1 12.25  per  share  ;  what 
is  his  commission,  at  £  per  cent.  ?  Ans.  $  3.36f . 

8.  My  agent,  S.  Cloon,  at  Cincinnati,  advises  me  that  he 
has  purchased  on  my  account  a  cargo  of  pork,  consisting  of 
700  barrels,  at  $  12.25  per  barrel ;  what  is  his  commission,  at 
If  per  cent.  ?  Ans.  $  150.06£. 

9.  Sent  to  my  agent,  John  Crowell,  at  Rochester,  N.  Y , 
$  8960,  to  purchase  a  quantity  of  flour ;  his  commissions  are  2 
per  cent,  on  the  purchase,  which  he  is  to  deduct  from  the  money 
sent  him  ;  what  is  his  commission  ?  Ans.  $175.68^f. 

10.  What  is  a  broker's  commission  on  the  sale  of  700  barrels 
of  flour,  at  $  5.75  per  barrel,  at  If  per  cent.  ?  Ans.  $  70.43f . 

11.  What  is  the  commission  on  the  sale  of  173cwt.  of  sugar, 
at  $  8.95  per  cwt.,  at  1£  per  cent.  ?  Ans.  $  29.03^. 

12.  My  agent  in  London  has  purchased  goods  for  me  to  the 
value  of  879  £.   12s.  9d  ;   what  is  his  commission,  at  3f  per 
cent.  ?  Ans.  29«£.  13s.  9tffod. 

13.  Sent  a  cargo  of  flour  to  Liverpool,  which  my  factor  sold 


SECT.  XLIV.]  STOCKS.  191 

for  987<£.  18s.  6d.     He  invested  this  sum  in  broadcloths,  at  !<£. 
3s.  8d.  per  yard.     His  commission   for  selling  the  flour  is 
per  cent.,  and  for  purchasing  the  broadcloth  1£  per  cent.,  an 
he  is  to  receive  his  commissions,  for  selling  and  buying,  out  of 
the  proceeds  of  the  flour.     Required  the  number  of  yards  of 
broadcloth  that  I  should  receive.  Ans.  800T4T8?%y<yyd- 

14.  I  have  remitted  to  my  correspondent  a  certain  sum  of 
money,  which  he  is  to  lay  out  for  me  in  iron,  and  having  re- 
served to  himself  2^-  per  cent,  on  the  purchase,  which  amount- 
ed to  $  90,  he  buys  for  me  the  iron,  at  $  95  per  ton.  Required 
the  sum  remitted,  and  the  quantity  of  iron  purchased. 

Sum  remitted,  $  3690. 

Iron  purchased,  37T.  17cwt  3qr. 


SECTION  XLIV. 
STOCKS. 

STOCKS  is  a  general  name  used  for  funds  established  by  gov- 
ernment, or  individuals,  in  their  corporate  capacity,  the  value  of 
which  is  often  variable. 

When  stocks  will  bring  more  in  the  market  than  their  origi- 
nal cost,  their  value  is  said  to  be  above,  par,  and  when,  from  any 
circumstance,  their  value  is  less  than  the  original  cost,  they  are 
said  to  be  below  par. 

The  method  for  computation  is  the  same  as  in  Interest. 

EXAMPLES. 

1.  What  is  the  value  of  $  24360  of  the  National  Bank  stock, 
at  135  per  cent.  ?  $  24360  x  1.35  —  $  32886  Ans. 

2.  Sold  15  shares,  $  100  each,  of  the  Boston  Bank,  at  13  per 
cent,  advance.     To  what  did  they  amount  ?         Ans.  $1695. 

3.  What  must  I  give  for  12  shares  in  the  Haverhill  Bank,  at 
15  per  cent,  advance,  shares  being  $100  each  ?    Ans.  $1380. 

4.  What  is  the  purchase  of  1058<£.  12s.  bank  stock,  at 
per  cent.  ?  Ans.  1225c£.  6s. 

5.  Sold  30  shares,  $100  each,  in  the  Boston  and  Providence 
Railroad,  at  8f-  per  cent,  advance.     To  what  did  they  amount  ? 

Ans.  $  3262.50. 

6.  What  is  the  value  of  10  shares  in  the  Philadelphia  and 
Trenton  Railroad  stock,  at  85  per  cent.,  original  shares  being 
$100?  Ans.  $850. 


192  INSURANCE.  [SECT.  XLV. 

7.  What  must  be  given  for  5  shares  of  the  stock  in  the  Ocean 
Insurance  Company,  at  7  per  cent,  advance,  the  original  shares 
being  $100?  Ans.  $535. 


SECTION  XLV. 
INSURANCE  AND  POLICIES. 

INSURANCE  is  a  security,  by  paying  a  certain  sum,  to  indem- 
nify the  secured  against  such  losses  as  shall  be  specified  in  the 
policy. 

POLICY  is  the  name  of  the  writ,  or  instrument,  by  which  the 
contract  or  indemnity  is  effected  between  the  parties. 

NOTE.  —  If  the  average  loss  does  not  exceed  5  per  cent,  the  underwrit- 
ers are  free,  and  the  insured  bears  the  loss  himself. 

RULE.  —  The  same  as  in  Interest  and  Discount. 

EXAMPLES. 

1 .  What  would  be  the  amount  of  the  insurance  on  my  house, 
valued  at  $  5728,  at  If  per  cent.  ?  Ans.  $100.24. 

2.  What  would  be  the  premium  for  insuring  my  good  ship 
Betsey,  for  a  voyage  from  Boston  to  Liverpool,  at  1£  per  cent., 
the  vessel  being  valued  at  $17,289.  Ans.  $  216.1 1£. 

3.  My  ship    Massachusetts  is  valued  at  $  50,765.     If  I  in- 
sure $10,000  at  the  Columbian  Office  at  4f  per  cent.,  $12,000 
at  the  Atlas  Office  at  3£,  and  the  vessel  is  cast  away  on  her 
voyage,  what  is  the  amount  of  my  loss  ?          Ans.  $  29,705. 

4.  A  gentleman  in  Boston  effected  an  insurance  on  his  store 
and  goods,  valued  at  $  47,600,  for  5  years.     For  the  first  year  he 
is  to  pay  4£  per  cent.,  for  the  second  year  3£  per  cent.,  for  the 
third  year  4f  per  cent,  for  the  fourth  year  5  per  cent.,  and  for 
the  fifth  year  5£  per  cent.     What  is  the  amount  of  his  whole 
insurance  ?  Ans.  811,007.50. 

5.  What  is  the  premium  on  $  1728  at  7^  per  cent  ? 

Ans.  $125.28. 

6.  My  ship,  the  Julia  Ann,  is  valued  at  $  35,000,  and  her 
cargo  at  $  75,000.     I  procure  an  insurance  on  -f  the  value  of 
the  ship,  at  3£  per  cent.,  and  on  f  of  her  cargo,  at  2£  per  cent. 
What  is  the  amount  of  premium  ?  Ans.  $1932.50. 

7.  The  good  ship  Marcia  Demming  is  valued  at  $18,750; 
her  cargo  cost  $  37,960,  and,  being  bound  to  Canton,  I  have  got 


SKCT.  XLVI.]  BANKING.  .  193 

$10,000  insured  on  the  vessel,  at  3f  per  cent.,  and  $  20,000  on 
her  cargo,  at  4£  per  cent.  What  would  be  my  loss  if  the  ves- 
sel should  founder  at  sea  ?  Ans.  $  27,997.50. 

8.  My  library  consists  of  2691  volumes,  valued  at  $3675. 
If  I  get  this  insured,  at  4£  per  cent,  what  is  my  actual  loss  if  it 
be  destroyed  ?  Ans.  $  179.15$. 

9.  What  is  the  premium  on  $  896,  at  12  per  cent.  ? 

Ans.  $107.52. 

10.  What  is  the  premium  on  $  850,  at  18f  per  cent.  ? 

Ans.  $157.25. 

11.  What  is  the  premium  for  insuring  $  9870,  at  7  per  cent.  ? 

Ans.  $  690.90. 

12.  What  sum  will  be  secured  for  a  policy  of  $  1728,  deduct- 
ing 15  per  cent.  ?  Ans.  $1468.80. 

13.  What  sum  must  a  policy  be  taken  out  for,  to  cover 
$  2475,  when  the  premium  is  10  per  cent.  ?       Ans.  $  2750. 

NOTE.  —  As  10  per  cent,  is  already  taken  out,  the  sum  covered  must  be 
I9o°0  of  the  policy. 

14.  A  certain   company  own  a  cotton   factory,  valued  at 
$  26,250.     For  what  sum  must  a  policy  be  taken  out  to  cover 
the  whole  property,  at  12£  per  cent.  ?    '  Ans.  $  30,000. 

15.  If  a  policy  be  taken  out  for  $  3600,  at  40  per  cent.,  what 
is  the  sum  covered  ?  Ans.  $2160. 

NOTE.  —  It  is  evident,  that,  if  40  per  cent,  be  taken  from  any  sum,  60 
per  cent,  will  be  left. 

16.  If  a  policy  be  taken  out  for  $  600,  at  10  per  cent.,  what 
is  the  sum  covered  ?  Ans.  $  540. 

17.  A  merchant  adventured  $  1000  from  Boston  to  New 
Orleans,  at  3  per  cent.  ;  from  thence  to  Chili,  at  5  per  cent.  ; 
from  thence  to  Canton,  at  6  per  cent. ;  and  from  thence  to 
Boston,  at  7  per  cent.     For  what  sum  must  he  take  out  a 
policy,  to  cover  his  adventure  the  voyage  round  ? 

Ans.  $1241.34,8+. 


SECTION  XL VL 
BANKING. 

A  BANK  is  a  place  of  deposit  for  money,  which  is  usually 
divided  into  shares,  and  owned  by  persons  called  stockholders. 
17 


194  BANKING.  [SECT.  XLVI. 

Its  concerns  are  managed  by  a  board  of  directors.  It  issues 
notes  or  bills  of  its  own,  intended  to  be  a  circulating  medium 
of  exchange  or  currency,  instead  of  gold  and  silver.  These 
bills  are  obtained  from  the  bank  by  loans.  When  money  is 
hired  from  a  bank,  it  is  the  usual  custom  to  deduct  the  interest 
at  the  time  of  receiving  it.  A  man,  therefore,  who  hires  money 
from  a  bank,  gives  his  note  for  a  sum  as  much  larger  than  he 
receives,  as  is  the  interest  of  the  note  for  the  given  time.  If  a 
man,  therefore,  gives  his  note  for  $100,  payable  in  63  days, 
he  receives  only  $  98.95. 

A  promissory  note  is  said  to  be  discounted,  when  it  is  re- 
ceived at  a  bank  as  a  security  for  money  taken  from  it ;  and 
the  interest  deducted  is  the  discount.  The  interest  on  every 
note  discounted  at  a  bank  is  computed  for  3  days  more  than 
the  time  specified  in  the  note  ;  that  is,  if  the  note  is  payable  in 
60  days,  the  interest  is  taken  for  63  days ;  for  the  law  allows 
3  days  to  the  debtor,  after  the  time  has  expired  for  payment, 
which  are  called  days  of  grace. 

The  rule  for  computing  the  discount  is  the  same  as  in  sim- 
ple interest. 

EXAMPLES. 

1.  What  is  the  bank  discount  on  $  476,  for  30  days  and 
grace?  Ans.  $2.61,8. 

2.  What  is  the  bank  discount  on  $1000,  for  60  days  and 
grace?  Ans.  $10.50. 

3.  What  is  the  bank  discount  on  $  7800,  for  90  days  and 
grace  ?  Ans.  $120.90. 

4.  What  is  the  bank  discount  on  $  8000,  for  60  days  and 
grace  ?  Ans.  $  84.00. 

5.  How  much    money  should   be  received   on  a  note  for 
$  760,  payable  in  5  months,  discounted  at  a  bank  when  the 
interest  is  6  per  cent.  ?  Ans.  $  740.62. 

6.  What  sum  is  paid  at  a  bank  for  a  note  of  $1728,  payable 
in  3  months  ?  Ans.  $  1 70 1 .2 1 ,6. 

7.  A  merchant  sold  a  cargo  of  hemp  for  $  7860,  for  which 
he  received  a  note  payable  in  6  months.     How  much  money 
will  he  receive  at  a  bank  for  this  note  ?          Ans.  $  7620.27. 

8.  A  merchant  bought  450  quintals  of  fish  at  $  3.50  cash, 
and  sold  them  immediately  for  $  4.00  on  6  months'  credit,  for 
which  he  received  a  note.   If  he  should  get  this  note  discounted 
at  a  bank,  what  will  he  gain  on  the  fish  ?          Ans.  $170.10. 


SECT.  XLYII.]  BARTER.  195 

SECTION  XLVII. 
BARTER/ 

BARTER  is  the  exchange  of  one  kind  of  merchandise  for 
another,  without  loss  to  either  party. 

Questions  in  this  rule  are  solved  by  finding  what  quantity 
of  goods,  at  a  given  price,  of  one  kind,  are  equal  in  value  to 
another  kind  of  goods  whose  price  is  also  given. 

EXAMPLES. 

1.  How  much  sugar,  at  12£  cents  per  lb.,  must  be  given  in 
barter  for  7601b.  of  raisins,  at  8  cents  per  lb.  ?       Ans.  486f  Ibs. 

2.  What  quantity  of  coffee,  at  17  cents  per  lb.,  must  be 
given  in  barter  for  7601b.  of  tea,  at  62£  cents  per  lb.  ? 

Ans.  2794T2Ylbs. 

3.  A  merchant  delivered  3  hogsheads  of  wine,  at  $1.10  per 
gal.,  for  126yd.  of  cloth  ;  what  was  the  cloth  per  yd.  ? 

Ans.  $1.65. 

4.  A  has  12cwt.  of  sugar,  worth  8  cents  per  lb.,  for  which 
B  gave  him  If  cwt.  of  cinnamon ;  at  what  did  B  value  his  cin- 
namon ?  Ans.-  $  0.54f  per  lb. 

5»  A  had  41cwt.  of  hops,  at  $  6.70  per  cwt.,  for  which  B 
gave  him  17cwt.  3qr.  41b.  of  prunes,  and  $  88  ;  what  were  the 
prunes  valued  at  per  lb.  ?  Ans.  $  0.09f  ££. 

6.  A  has  sugar  which  he  barters  with  B,  for  4  cents  per  lb. 
more  than  it  cost  him,  against  tea  which  cost  B  40  cents  per 
lb.,  but  which  he  puts  in  barter  at  50  cents.     What  did  A's 
sugar  cost  him  per  lb.  ?  Ans.  $  0.16. 

7.  How  many  staves,  at  $  25  per  thousand,  must  a  merchant 
receive  for  15  hogsheads  of  wine,  at  $1.25  per  gallon  ? 

Ans.  47£M.  staves. 

8.  Q  has  670  bushels  of  oats,  which  cost  him  35  cents  per 
bushel ;  these  he  barters  with  Z,  at  50  cents  per  bushel,  for 
flour  that  cost  Z  $  5.00  per  barrel.    What  is  the  bartering  price 
of  the  flour,  and  how  much  will  Q  receive  ? 

Ans.  46^  barrels  of  flour,  at  $7.14f  per  barrel. 

9.  S.  Jenkins  has  73T9^9T  bushels  of  corn,  which  is  worth  7s. 
per  bushel ;  but  in  barter  he  is  willing  to  put  it  at  6s.  8d.,  pro- 
vided he  can  have  wheat  worth  7s.  6d.  per  bushel  for  7s.  3d. 
Will  he  gain  or  lose,  and  how  much  per  cent.  ? 

Ans.  Lose  1^  per  cent. 


196 


PRACTICE. 


[SECT.  XLVIII. 


SECTION  XLVIII. 


PRACTICE. 

PRACTICE  is  an  expeditious  way  of  performing  questions  in 
Compound  Multiplication  and  Proportion. 

RULE.  —  Assume  the  price  at  some  unit  higher  than  the  given  price; 
that  is,  if  the  price  be  pence,  or  pence  and  farthings,  assume  the  price 
at  a  shilling  a  yard,  or  pound,  <%c. ;  if  the  price  be  in  shillings,  or 
shillings  and  pence,  <£c.,  assume  the  price  at  a  pound  a  yard,  <5fC. ; 
then  take  the  aliquot  parts  of  a  pound. 

TABLE, 

Showing  the  aliquot  parts  of  Money  and  Weights. 


Parts  of  a  £. 

Parts  of  a  ton. 

Parts  of  a  half-cwt. 

a.       d. 

cwt.     qr. 

Ib. 

10          is    £ 

10          is     £ 

28    is      £ 

6    8    «     £ 

5          "     J 

14    "      ^ 

5           "     ± 

4          "     i 

8    "      | 

4          «    £ 

2    2    "     i 

7    *'     4- 

3    4    "     $ 

2           *   TV 

4    "     TX 

2    6    "     i 

3^  "     TV 

2          "    -iV 

2     "     2-V 

i  s  ;;  A 

Parts  of  a  cwt. 

Parts  of  a  shilling. 

qr.      Ib. 
2            is     £ 

Parts  of  a  quarter-cwt. 

d. 

1                "       J- 

Ib. 

6        is        £ 

16    "     ^ 

14    is      $ 

4        «        4 

14    "     i 

7     "      i 

3        "        ± 

8      **      TV 

^      4     "      ,f 

2        "         £ 

7    «    T^. 

31  tt       i 

1£      "        I 

4"      i 

mm 

2     "     A 

1            "          TV 

2    "     3V 

i   "   A 

EXAMPLES. 


1.    What   will   368   yards   of  ribbon   cost,   at   6   pence   a 
yard  ?  Ans.  9<£.  4s. 


SECT.  XLVIII.] 


PRACTICE. 


197 


OPERATION. 


6d.  i= 


20)184 


We  assume  the  price  at  a  shilling 
a  yard,  and  then  say,  if  368  shillings 
be  the  price  at  a  shilling  a  yard,  at 
6  pence  it  must  be  half  as  much, 
viz.  184  shillings.  We  then  reduce 
the  shillings  to  pounds. 


cost,  at 
Ans. 


8 


pence 
.  10s. 


Having  found  the 
price  at  6d.  as  be- 
fore, we  find  it  for 
the  2d.  by  saying 
that  2d.  is  of  6d. 


2.  What  will  4785  yards  of  cotton 
yard? 

OPERATION. 

6d.  =  £)4785s.  =  price  at  1  shilling. 
2d.  =  £)2392s.  6d.  =  price  at  6d. 
797s.  6d.  =  price  at  2d. 

20)3190s.  Od. 

159<£.  10s.  =  price  at  8d. 

3.  What  is  the  interest  of  $  368,  at  15  per  cent.  ? 

Ans.  $  55.20. 

OPERATION. 

10  per  cent.  =  TV)368 

5  —  £)36.80         10  per  cent,  is  -fa  of  the  principal. 

18.40    an(j  5  per  cent.  is  £  of  10  per  cent. 
$  55.20 

4.  What  is  the  value  of  17  acres  3  roods  35  rods  of  land, 
at  $  80  per  acre  ?  Ans.  $  1437.50. 


2R.  =  £ 
1R.  SB;} 

20rd.  =  | 
lOrd.  SB,  J 

5rd.  = 


OPERATION. 

$80 
17 
560 
80 

$1360  =  price  of  17  A. 
£)     40  =  do.  of  2R. 
20  =  do.  of  1R. 
10  =  do.  of  20rd. 
5  =  do.  of  lOrd. 
2.50  =  do.  of  5rd. 


By  dividing  the  price  of 
1  acre  by  2,  we  obtain  the 
price  of  2R.  ;  and  by  halv- 
ing this,  we  find  the  price 
of  1R.  ;  and  as  20  rods  is 
half  of  a  rood,  its  value  will 
be  one  half;  and  in  the 
same  manner  10  rods  will 
be  half  the  price  of  20  rods, 
and  5  rods  will  be  half  the 
price  of  10  rods. 


1437.50  =  price  of  17A.  3R.  35rd. 


5.  What  cost  14  tons  15cwt.  3qr.  211b.  of  iron,  at  $  120  per 
ton?  Ans.  $1775.62,5. 

17* 


198  PRACTICE.  [SECT.  XLVIII. 

OPERATION. 

$120 
14 

$1&?0.00  =  price  of  14T. 
lOcwt.  =  £)     60.00  =  do.  of  lOcwt. 

$1740.00 

5cwt.  =  £)     30.00  =  do.  of  5cwt. 

2qr.  =  ,U       3.00  =  do.  of  2qr. 

Iqr.  =  £)       1.50  =  do.  of  Iqr. 

141b.  =  I)         .75  =  do.  of  14lb. 

Tib.  =  j)         .375  =  do.  of  Tib. 

$1775.62,5=:do.  of  14T.  15cwt.  3qr.211b. 

6.  What  cost  3871b.  of  sugar,  at  9  pence  a  pound  ? 

Ans.  14^.  10s.  3d. 

7.  What  cost  4981b.  of  green  tea,  at  2  shillings  and  6  pence 
per  pound  ?  Ans.  62 £.  5s.  Od. 

8.  What  cost  384  yards  of  cloth,  at  4  shillings  and  9  pence 
a  yard?  Ans.  91  £.  4s.  Od. 

9.  What  cost  714  yards  of  broadcloth,  at  15  shillings  and  6 
pence  per  yard  ?  Ans.  553«£.  7s.  Od. 

10.  What  cost  16cwt.  3qr.  lOlb.  of  copperas,  at  $  2.50  per 
cwt  ?  Ans.  $  42.09,8. 

11.  What  cost  27cwt.  Iqr.  211b.  of  coffee,  at  $14  per  cwt.  ? 

Ans.  $384.12£. 

12.  What  cost  7  tons  13cwt.  2qr.  Tib.  of  hay,  at  $  24.60  per 
ton?  Ans.  $188.88,1$. 

13.  If  1  acre  of  land  cost  $  80.50,  what  will  25  acres  2  roods 
35  rods  cost  ?  Ans.  $  2070.35,9 1. 

14.  If  1  acre  cost  $  32.32,  what  will  51  A.  OR.  lord,  cost  ? 

Ans.  $1651.35. 

15.  If  1  yard  of  cloth  cost  $  5.60,  what  will  7yd.  3qr.  2na. 
cost?  Ans.  $44.10. 

16.  What  is  the  premium  on  $  6780,  at  12£  per  cent.  ? 

Ans.  $  847.50. 

17.  What  is  the  interest  of  $1728  for  5  years  7  months  and 
20  days  ?  Ans.  $  584.64. 

18.  What  will  19  tons  19cwt.  3qr.  27^-lb.  of  copperas  cost, 
at  19<£.  19s.  llf d.  per  ton  ?  Ans.  399c£.  19s.  5|f f^d. 

19.  The  estimated  distance  of  a  certain  railroad  is  14m.  3fur. 
35rd.  10ft  ;  what  would  be  the  expense  of  constructing  it,  at 
$  18675  per  mile  ?  Ans.  $  270531.07+. 


SECT.  XLIX.]  EQUATION  OF  PAYMENTS.  199 

SECTION  XLIX. 
EQUATION  OF  PAYMENTS. 

WHEN  several  sums  of  money,  to  be  paid  at  different  times, 
are  reduced  to  a  mean  time  for  the  payment  of  the  whole,  with- 
out gain  or  loss  to  the  debtor  or  creditor,  it  is  called  Equation 
of  Payments. 

EXAMPLES. 

1.  A  owes  B  $19,  $5  of  which  is  to  be  paid  in  6  months, 
$  6  in  7  months,  and  $  8  in  10  months.     What  is  the  medium 
time  for  the  payment  of  the  whole  ? 

OPERATION.  By  analysis.     $5  for  6  months 

P  5  X    6  =  30  is  the  same  as  $1   for  30  months  ; 

and  $  6  for  7  months  is  the  same 

as  $1  for  42  months  ;  and  $  8  for 

19         19)152(8  months.     10  months  is  the  same  as  $1  for  80 

152  months  ;  therefore  $1  for  30  +  42 

-f-  80  =  152  months  is  the  same 

as  $  5  for  6  months,  $  6  for  7  months,  and  $  8  for  10  months ; 
but  $  5,  $  6,  and  $  8  are  $19  ;  therefore  $1  for  152  months  is 
the  same  as  $19  for  y1^  of  152  months,  which  is  8  months,  as 
before.  Hence  the  propriety  of  the  following 

RULE.*  —  Multiply  each  payment  by  the  time  at  which  it  is  due ;  then 
divide  the  sum  of  the  products  by  the  sum  of  the  payments,  and  the  quo- 
tient wiU  be  the  true  time  required. 

2.  A  owes  B  $  300,  of  which  $  50  is  to  be  paid  in  2  months, 
$100  in  5  months,  and  the  remainder  in  8  months.     What  is 
the  equated  time  for  the  whole  sum  ?  Ans.  6  months. 

*  This  is  the  rule  usually  adopted  by  merchants,  but  it  is  not  perfectly 
correct ;  for  if  I  owe  a  man  $>  200,  $100  of  which  I  was  to  pay  down,  and 
the  other  $100  in  two  years,  the  equated  time  for  the  payment  of  both 
sums  would  be  one  year.  It  is  evident,  that,  for  deferring  the  payment  of 
the  first  $  100  for  1  year,  I  ought  to  pay  the  amount  of  $  100  for  that  time, 
which  is  $  1 06 ;  but  for  the  other  $  100,  which  I  pay  a  year  before  it  is 
due,  I  ought  to  pay  tine  present  worth  of  $  100,  which  is  $94.33fg,  where- 
as, by  Equation  of  Payments,  I  only  pay  $200.  Strict  justice  would  there- 
fore demand  that  interest  should  be  required  on  all  sums  from  the  time 
they  become  due  until  the  time  of  payment,  and  the  present  worth  of  all 
sums  paid  before  they  are  due.  The  better  rule  would  be,  to  find  the  pres- 
ent worth  on  each  of  the  sums  due,  and  then  find  in  what  time  the  sum 
of  these  present  worths  would  amount  to  the  payments. 


200  EQUATION   OF  PAYMENTS.  [SECT.  XLIX. 

3.  There  is  owing  to  a  merchant  $  1000  ;  $  200  of  it  is  to  be 
paid  in  3  months,  $  300  in  5  months,  and  the  remainder  in  10 
months.     What  is  the  equated   time  for  the  payment  of  the 
whole  sum  ?  Ans.  7  months  3  days. 

4.  A  owes  B  $150,  $  50  to  be  paid  in  4  months,  and 

in  8  months.  B  owes  A  $  250  to  be  paid  in  10  months.  It  is 
agreed  between  them  that  A  shall  make  present  payment  of  his 
whole  debt,  and  that  B  shall  pay  his  so  much  sooner  as  to  bal- 
ance the  favor.  I  demand  the  time  at  which  B  must  pay  the 
$  250.  Ans.  6  months. 

5.  A  merchant  has  $144  due  him,  to  be  paid  in  7  months, 
but  the  debtor  agrees  to  pay  one  half  ready  money,  and  one 
third  in  4  months.     What  time  should  be  allowed  him  to  pay 
the  remainder  ?  Ans.  2  years  10  months. 

6.  There  is  due  to  a  merchant  $  800,  one  sixth  of  which  is 
to  be  paid  in  2  months,  one  third  in  3  months,  and  the  remain- 
der in  six  months  ;  but  the  debtor  agrees  to  pay  one  half  down. 
How  long  may  the  debtor  retain  the  other  half  so  that  neither 
party  may  sustain  loss  ?  Ans.  8§  months. 

7.  I  have  purchased  goods  of  A.  B.  at  sundry  times  and  on 
various  terms  of  credit,  as  by  the  statement  annexed.     When 
is  the  medium  time  of  payment  ? 

Jan.        1,  a  bill  amounting  to  $  375.50  on  4  months'  credit. 
"        20,      "  "  168.75  on  5  months'  credit. 

Feb.       4,       u  "  386.25  on  4  months'  credit. 

March  11,       "  "  144.60  on  5  months' credit. 

April     7,       "  386.90  on  3  months' credit. 

FORM   OF   STATEMENT. 

Due  May    1,  $375.50 

June  20,  168.75  x  50=  843750 
June  4,  386.25  x  34  =  1313250 
Aug.  11,  144.60  x  102  =  1474920 
July  7,  386.90  x  67  =  2592230 

$1462.00 )6224150(42£f££da.  Ans. 

584800 

376150 
292400 

83750 

The  medium  time  of  payment  will  therefore  be  42J-f£f  days, 
that  is,  43  days  from  May  1,  which  will  be  June  12. 


«ECT.  i.]  CUSTOM-HOUSE  BUSINESS.  201 

S.  I  have  sold  to  C.  D.  several  parcels  of  goods,  at  sundry 
times,  and  on  various  terms  of  credit,  as  by  the  statement  an- 
nexed. 

Jan.        1,  a  bill  amounting  to  $  600  on  4  months'  credit 

Feb.       7,       «             "  370  on  5  months' credit. 

March  15,       "             "  560  on  4  months'  credit. 

April    20,       "             "  420  on  6  months'  credit. 

When  is  the  equated  time  for  the  payment  of  all  the  bills  ? 

Ans.  July  11. 

9.  Purchased  goods  of  John  Brown,  at  sundry  times,  and  on 
various  terms  of  credit,  as  by  the  statement  annexed. 

March     1,  1845,  a  bill  amounting  to  $  675.25  on  3  months. 
July       4,      "  «  "    '  376. 18  on  4  months. 

Sept.     25,      "  "  "  821.75  on  2  months. 

Oct.        1,     "  "  »  961.25  on  8  months. 

Jan.        1,   1846,       «  "  144.50  on  3  months. 

Feb.      10,     "  "  "         •      81 1.30  on  6  months. 

March  12,      "  «  «  567.70  on  5  months. 

April    15,      "  369.80  on  4  months. 

What  is  the  equated  time  for  the  payment  of  the  above  bills  ? 

Ans.  March  16,  1846. 


SECTION  L. 
CUSTOM-HOUSE   BUSINESS. 

IN  every  port  of  the  United  States  where  merchandise  is 
either  exported  or  imported,  there  is  an  establishment  called  a 
Custom-house.  Connected  with  this  are  certain  officers,  ap- 
pointed by  government,  called  custom-house  officers,  whose 
business  is  to  collect  the  duties  on  various  kinds  of  merchandise, 
&c.,  imported  into  the  United  States. 

The  following  article  on  Allowances,  &c.,  was  very  politely 
furnished  the  author  by  the  officers  of  the  Boston  custom-house, 
and  may  therefore  be  relied  on  as  perfectly  correct. 

Allowances. 

Draft,  is  an  allowance  made  by  the  officers  of  the  United 
States  government  in  the  collection  of  duties  on  merchandise 
liable  to  a  specific  duty,  and  ascertained  by  weight,  and  is  also 


202  CUSTOM-HOUSE  BUSINESS.  [SECT.  L. 

given  by  the  usage  of  merchants  in  buying  and  selling.  It  is  a 
deduction  from  the  actual  gross  weight  of  the  article  paying  du- 
ty by  the  pound  or  sold  by  weight. 

For  example,  a  box  of  sugar  actually  weighs  500  pounds. 
The  draft  upon  this  weight  is  4  pounds. 

500  gross. 
4  draft. 

496  difference.  Upon  this  difference  is  made  a  further  allow- 
ance of  fifteen  per  cent,  as  tare,  or  as  the  actual  weight  of  the 
box  before  the  sugar  was  put  into  it  This  tare  is  allowed  by 
the  government  in  the  collection  of  the  duty,  and  by  the  mer- 
chant in  buying  and  selling.  Take,  then,  the  box  of  sugar,  say 

5001b.  gross. 
4  draft. 

496"  difference. 
74  tare. 

422  net  weight,  upon  which  a  duty  is  paid  to  the  government, 
or  price  is  paid  to  the  merchant  in  his  sale.  This  tare  of  74 
pounds,  or  15  per  cent.,  is  usually  more  than  the  actual  tare,  but 
is  assumed  as  the  probable  or  actual  tare,  by  reason  of  the  im- 
possibility of  "  starting  "  every  box  to  ascertain  the  actual  weight 
of  the  sugar,  and  the  actual  weight  of  the  box  which  contains  it. 
This  tare  is  sufficiently  correct  for  the  collector  of  the  duty, 
and  the  merchant  who  deals  in  the  article.  It  is  intended  to 
be  a  liberal  allowance,  and  varies  but  little  from  the  actual 
tare. 

Drafts  allowed  at  the  custom-house  in  the  collection  of  du- 
ties, and  by  the  merchants  in  their  purchases  and  sales,  are  as 
follow  :  — 

Allowance  for  Draft. 

Draft,  is  another  name  for  Tret^  which  is  an  allowance  in 
weight  for  waste. 

lb.  lb. 

On  112  1 

Above  112  and  not  exceeding  224  2 

"      224         "  "          336  3 

"      336         "  "        1120  4 

"    1120         "  "        2016  7 

«    2016  9 

EXPLANATION.  ^—  Many  articles  of  merchandise  are  weighed 
separately ;  for  example,  boxes  and  casks  of  sugar,  chests  of 


SECT.  L.]  CUSTOM-HOUSE  BUSINESS.  203 

tea  and  indigo.  Upon  each  box  or  cask,  or  chest,  an  allowance 
should  be  made  for  draft,  according  to  its  weight,  as  by  the 
above  rule.  Bags  of  sugar  and  coffee,  or  bars  of  iron  and  bun- 
dles of  steel,  might  be  weighed  together ;  say,  10  bags  of  coffee 
at  one  draft  might  weigh  1121  pounds  ;  from  this  gross  weight 
must  be  deducted  7  pounds  as  draft ;  35  bars  of  Russia  iron 
might  be  weighed  at  one  draft,  —  weight  2250  pounds,  upon 
which  would  be  an  allowance  of  9  pounds  draft,  and  by  law 
and  usage  there  can  be  no  greater  allowance  than  9  pounds  for 
draft.  A  greater  or  less  number  of  bags  of  coffee,  or  bars  of 
iron,  or  any  other  article  of  merchandise,  is  weighed,  and  the 
deduction  is  according  to  the  weight  of  each  draft.  An  old  rule, 
and  probably  a  better  one,  among  merchants  was  the  allowance 
of  £  per  cent,  on  the  gross  weight  of  all  merchandise  weighed, 
as  draft. 

Allowance  for  Leakage. 

Two  per  cent,  is  allowed  on  the  gauge  of  ale,  beer,  porter, 
brandy,  gin,  molasses,  oil,  wine,  and  rum,  and  other  liquors  in 
casks,  besides  the  real  wants  of  the  cask  ;  for  example,  a 
cask  of  molasses  may  gauge  140  gallons,  gross  gauge  ;  from 
this  first  deduct  5  gallons,  the  actual  wants,  or  the  quantity  ne- 
cessary to  fill  the  cask,  —  we  have 

140  gross. 

5  out. 
135  difference. 

3  two  per  cent,  for  leakage. 
132  gallons  net. 

Tare  is  an  allowance  made  for  the  actual  or  supposed  weight 
of  the  cask,  box,  case,  or  bag,  which  contains  the  article  of 
merchandise. 

The  usage  of  merchants  is  in  conformity  with  the  law  and 
usage  of  the  officers  of  the  customs  in  their  allowance  for  tare, 
directed  by  law,  or  found  to  be  correct  by  their  examination  and 
experience. 

The  tariff  of  the  United  States  being  in  its  details  so  unset- 
tled, it  is  deemed  advisable  not  to  insert  any  table. 

EXAMPLES. 

1.  Find  the  net  weight  of  a  hogshead  of  sugar,  weighing 
gross  12281b.,  tare  12  per  cent.  Ans.  10751b. 


204  CUSTOM-HOUSE   BUSINESS.  [BKCT.  i. 

OPERATION. 

12281b.  gross  weight. 
JTlb.  draft. 

1221 
12  per  cent,  of  122 lib.  —  1461b.  tare. 

10751b.  net  weight. 
NOTE.  —  For  draft  let  the  pupil  examine  page  202. 

2.  Required  the  net  weight  of  6  boxes  of  sugar,  weighing 
gross  as  follows,  the  tare  being  15  per  cent.  :  - — 

No.  1,  450  Ibs. 

No.  2,  470  do.  OPERATION. 

No.  3,  510  do.  29141b.  gross. 

No.  4,  496  do.  6x4  =     241b.  draft 

No.  5,  468  do.  2890 

No.  6,520  do.      Tare  15  per  cent.  433 
Gross,  2914  24571b.  net  weight. 

3.  What  is  the  net  weight  of  4  chests  of  tea,  which  weigh  as 
follows,  tare  22  per  cent.  ?  Ans.  3841b. 

OPERATION. 

No.  1,  120  Ibs.  4801b.  gross. 

No.  2,  116  do.  81b.  draft. 

No.  3,  126  do.  472 

No.  4,  US  do.  22  x  4  =_881b.  tare. 
480  do.  3841b.  net. 

OPERATION. 

4.  What  is  the  net  weight  of  5  bags                   53®}£' 
of  pepper,  weighing  as  follows  :  1081b., 

1  laib.,  lOOlb.,  120lb.,  and  981b.,  tare  2  535 

per  cent.  ?  Ans.  524.       2  Per  cent^lllb.  tare. 

5241b.  net. 

OPERATION. 

5.  What  is  the  net  weight  of  4 

,                              ...           °t>  ,,                                         51b.  arait. 
kegs  of  mace,  weighing  as  follows  :  

1121b.,  1201b.,  1181b.,  and  llOlb., 

tare  33  per  cent.  ?      Ans.  3051b.         33  Per  cent-  15Qlb-  **Te. 

3051b.  net. 

NOTE.  —  In  making  allowances,  if  there  be  a  fraction  of  more  than  half 
a  pound,  1  pound  is  added  to  the  tare. 


IECT.    LI.]  RATIO.  205 

AMERICAN  DUTIES. 

The  duties  on  merchandise  imported  into  the  United  States 
are  either  specific  or  ad  valorem  duties. 

Specific  duty  is  a  certain  sum  paid  on  a  ton,  hundred  weight, 
pound,  square  yard,  gallon,  &c. ;  but  when  the  duty  is  a  certain 
per  cent,  on  the  actual  cost  of  the  goods  in  the  country  from 
which  they  are  imported,  it  is  called  an  ad  valorem  duty,  that 
is,  a  duty  according  to  the  value  of  the  article. 

6.  What  is  the  duty  on  6  hogsheads  of  sugar,  weighing 
gross   as   follows:    No.    1,    1276lb.,  No.   2,   12801b.,  No.   3, 
11781b.,  No.  4,  13781b.,  No.  5,  15701b.,  No.  6,  1338lb.  ;  duty 
2J-  cents  per  lb.,  tare  12  per  cent.  ?  Ans.  $175.52,5. 

7.  What  is  the  duty  on  an  invoice  of  woollen  goods,  which 
cost  in  London  £  986  sterling,  at  44  per  cent,  ad  valorem,  the 
pound  sterling  being  $  4.84  ?  Ans.  $  2099  78-|-. 

8.  Required  the  duty  on  5  pipes  of  Port  wine,  gross  gauge 
as  follows  :  No.  1,  176  gallons,  No.  2,  145  gallons,  No.  3,  128 
gallons,  No  4,  148  gallons,  No.  5,  150  gallons  ;  wants  of  each 
pipe,  4  gallons  ;  duty  15  cents  per  gallon.        Ans.  $106.80. 

9.  Required  the  duty  on  a  cargo  of  iron,  weighing  270  tons, 
at  •$  30  per  ton  ?  Ans.  $  8100. 

10.  Compute  the  duty  on  7890  pounds  of  tarred  cordage,  at 
4  cents  per  pound  ;  duty  If  per  cent.  Ans.  $310.08. 

11.  What  duty  should  be  paid  on  10  casks  of  nails,  weighing 
each  4501b.  gross,  at  4  cents  per  lb.  ?  Ans.  $164.12. 


SECTION  LI. 

RATIO. 

RATIO  is  the  relation  which  one  quantity  bears  to  another  of 
the  same  kind  with  respect  to  magnitude  ;  and  the  comparison 
is  made  by  considering  how  often  the  one  is  contained  in  the 
other,  or  how  often  the  one  contains  the  other.  Thus,  the  ratio 
of  12  to  3  is  expressed  by  dividing  12  by  3  —  -\£  =  4,  ratio  ; 
or  it  may  be  expressed  by  dividing  3  by  12  =  T32-  =  ^,  ratio. 
The  former  is  the  method  by  which  the  English  mathematicians 
express  ratio,  and  the  latter  is  the  French  method. 

The  former  of  these  quantities  is  called  the  antecedent,  and 
the  latter  the  consequent. 
18 


206  RATIO.  [SECT.  LI. 

When  the  antecedent  is  equal  to  the  consequent,  it  is  called 
a  ratio  of  equality  ;  thus  the  ratio  of  6  to  6  =  f  =  1.  But  if 
the  antecedent  be  larger  than  the  consequent,  it  is  a  ratio  of 
greater  inequality  ;  and  if  the  antecedent  be  less  than  the  con- 
sequent, it  is  a  ratio  of  less  inequality. 

The  antecedent  and  consequent  are  called  the  terms  of  the 
ratio  ;  and  the  quotient  of  the  two  terms  is  the  index  or  expo- 
nent of  the  ratio. 

Compound  ratio  is  made  up  of  two  or  more  ratios,  by  multi- 
plying their  terms  and  exponents  together. 

The  ratio  of  8  to  6  and  of  4  to  2  may  be  compounded ;  thus, 
8  to  6  =  |;  4  to  2  —  £  ; 
8  x  4  to  6  x  2  =  g| ;  32  to  12  =  f  } , 

If  a  ratio  be  compounded  of  two  equal  ratios,  it  is  called  a 
duplicate  ratio  ;  of  three  ratios,  it  is  called  a  triplicate  ratio,  dec. 

Thus,  if  the  ratio  of  4  to  2  be  2,  and  the  ratio  of  6  to  3  be 
2,  the  ratio  of  4  X  6  to  2  X  6  will  be  2  X  2,  that  is,  the  ratio 
of  24  to  12  will  be  22,  &c. 

If  the  terms  of  a  ratio  be  prime  to  each  other,  no  quantities 
can  be  found  in  the  same  ratio  but  what  would  be  multiples 
thereof. 

Numbers  that  are  prime  to  each  other  are  the  least  of  all 
numbers  in  the  same  ratio. 

If,  therefore,  we  wish  to  ascertain  whether  the  ratio  of  3  to 
7  is  greater  or  less  than  the  ratio  of  4  to  9,  since  these  ratios 
are  represented  by  the  fractions  y  and  $-,  we  reduce  them  to  a 
common  denominator,  f  £  and  f  f  ;  and,  since  the  latter  of  these 
is  greater  than  the  former,  it  is  evident  that  the  ratio  of  3  to  7 
is  less  than  the  ratio  of  4  to  9. 

If  we  have  the  terms  of  a  ratio  given  in  large  numbers,  that 
are  prime  to  each  other,  and  we  wish  to  find  a  ratio  nearly 
equivalent,  whose  terms  are  expressed  by  smaller  numbers,  we 
adopt  the  following 

RULE.  —  Divide  the  greater  term  by  the  Jess,  and  that  divisor  by  the 
remainder,  as  in  Sect.  XVI. ,  Case  I.,  of  Vulgar  Fractions.  Then,  if 
the  antecedent  be  greater  than  the  consequent,  the  first  quotient  divided  by 
1  gives  the  first  ratio  ;  if  less,  a  unit  divided  by  the  first  quotient  will 
express  the  first  ratio. 

Multiply  the  terms  of  the  first  ratio  by  the  second  quotient,  and  add  a 
unit  to  the  numerator  or  denominator,  according  as  the  antecedent  of 
the  original  terms  is  greater  or  less  than  its  consequent,  and  ice  have 
the  second  ratio. 

Then,  as  a  general  principle,  we  multiply  fhe  terms  of  the  ratio  last 


SECT.  LII.J  PROPORTION.  207 

found  by  tlie  next  succeeding  quotient,  and  to  the  product  we  add  the  cor- 
responding terms  of  the  preceding  ratio,  and  we  have  the  next  succeed- 
ing ratio ;  and  thus  we  proceed  until  there  is  no  remainder,  or  until  we 
have  arrived  at  a  sufficient  approximation. 

1.  Let  it  be  required  to  find  a  series  of  ratios  in  less  num- 
bers, constantly  approaching  to  the  ratio  of  314159  to  100000, 
which  is  nearly  the  ratio  of  the  circumference  of-  a  circle  to  its 
diameter. 

OPERATION. 

100000)314159(3 

300000   ^ 
14159)100000(7 
99113 

887~)  14159(15 
13305 

"854)887(1 
854 

33,  &c. 
3  =  f ,  the  first  ratio. 

f3*7)+1  =  2T2-,  the  second  ratio,  being  the  approximation  of  Ar- 

[chimedes. 
third  ratio. 

fourth  ratio,  the  approximation  of  Metius. 

-f-  •}•,  &c.,  in  a  continued  frac- 
tion. 


SECTION  LII. 

PROPORTION. 

PROPORTION  is  the  likeness  or  equalities  of  ratios.  Thus,  be- 
cause 5  has  the  same  relation  or  ratio  to  10  that  8  has  to  16, 
we  say  such  numbers  are  in  proportion  to  each  other,  and  are 
therefore  called  proportionals. 

If  any  four  numbers  whatever  be  taken,  the  first  is  said  to 
have  the  same  ratio  or  relation  to  the  second,  thakthe  third  has 
to  the  fourth,  when  the  first  number  or  term  contains  the  second 


208  PROPORTION.  [SECT.  LII. 

as  many  times  as  the  third  contains  the  fourth,  or  when  the 
second  contains  the  first  as  many  times  as  the  fourth  does  the 
third.  Thus,  8  has  the  same  ratio  to  4  that  12  has  to  6,  be- 
cause 8  contains  4  as  many  times  as  12  does  6.  And  3  has  the 
same  relation  to  9  that  4  has  to  12,  because  9  contains  3  as 
many  times  as  12  does  4.  Ratios  are  represented  by  colons, 
and  the  equalities  of  ratios  by  double  colons. 

3  :  9  : :  8  :  24  is  read  thus  :  —  3  has  the  same  ratio  or  relation 
to  9  as  8  to  24.  The  first  and  third  numbers  of  a  proportion 
are  called  antecedents,  and  the  second  and  fourth  are  called 
consequents ;  also,  the  first  and  fourth  are  called  extremes,  and 
the  second  and  third  are  called  means. 

Whatever  four  numbers  are  proportionals,  if  their  antece- 
dents or  consequents  be  multiplied  or  divided  by  the  same  num- 
bers, they  are  still  proportionals  ;  and  if  the  terms  of  one  pro- 
portion be  multiplied  or  divided  by  the  corresponding  term  of 
another  proportion,  their  products  and  quotients  are  still  propor- 
tionals. 

This  will  appear  evident  from  the  various  changes  that  the 
following  example  admits. 

4  :  8  : :  3  :  6  Directly. 

8  :  4  : :  6  :  3  By  inversion. 

4  :  3  : :  8  :  6  By  permutation. 
4-}-8:8::3-f-6:6By  composition. 
4:4-f8::3:3-|-6By  composition. 
4:8  —  4::3:6  —  3  By  division. 
8  —  4:8::  6  —  3:6  By  division. 
4X4:8x8::3x3:6x6By  compound  ratios. 

£  :  f  : :  £  :  f  By  division. 

That  the  product  of  the  extremes  is  equal  to  that  of  the 
means  is  evident  from  the  following  consideration.  Let  the  fol- 
lowing proportionals  be  taken.  12  :  3  : :  8  :  2.  From  the  def- 
inition of  proportion,  the  first  term  contains  the  second  as  many 
times  as  the  third  does  the  fourth  ;  therefore,  -^  —  f ;  but  -^  = 
*gt,  and  J  =  ^t ;  and  if  24,  the  numerator  of  the  first  fraction, 
which  is  a  substitute  for  the  first  term,  be  multiplied  by  6,  the 
denominator  of  the  second  fraction,  and  a  substitute  for  the 
fourth  term,  the  product  will  be  the  same  as  if  6,  the  denomi- 
nator of  the  first  fraction,  and  a  substitute  for  the  second  term, 
be  multiplied  by  24,  the  numerator  of  the  second  fraction,  and 
a  substitute  fot  the  third  term.  Thus  24  x  6  =  6  X  24.  There- 
fore the  product  of  the  extremes  is,  in  all  cases,  equal  to  that 
of  the  means. 


SECT,  ni.]  PROPORTION.  209 

If,  then,  one  of  the  extremes  be  wanting,  divide  the  product 
of  the  means  by  the  extreme  given ;  or,  if  one  of  the  means 
be  wanting,  divide  the  product  of  the  extremes  by  the  means 
given,  and  the  result  will  be  the  term  sought. 

To  apply  this,  we  will  take  the  following  question.  If  5 
yards  of  cloth  cost  $15,  what  will  7  yards  cost  ?  It  is  evident 
that  twice  the  quantity  of  cloth  would  cost  twice  the  sum,  and 
that  three  times  the  quantity,  three  times  the  sum,  &c.  ;  that 
is,  the  price  will  be  in  proportion  to  the  quantity  purchased. 
We  then  have  'three  terms  of  a  proportion  given,  one  of  the 
extremes  and  the  two  means,  to  find  the  other  extreme. 

Thus,  5:7::  15.  Therefore,  to  find  the  other  extreme  by 
the  rule  above  stated,  we  multiply  the  two  means,  7  and  15,  and 
divide  their  product  by  the  extreme  given,  and  the  quotient  is 
the  extreme  required.  7  X  15  =  105.  105  -H  5  =  21  dollars, 
the  answer  required. 

To  perform  this  question  by  analysis,  we  reason  thus.  If  5 
yards  cost  15  dollars,  1  yard  will  cost  one  fifth  as  much,  which 
is  3  dollars  ;  and  if  1  yard  cost  3  dollars,  7  yards  will  cost  7 
times  as  much,  which  is  21  dollars. 

RULE.*  —  State  the  question  by  making  that  number  which  is  of  the 
same  name  or  quality  of  the  answer  required  the  third  term;  then,  if 
the  answer  required  is  to  be  greater  than  the  third  term,  make  the  second 
term  greater  tftan  the  first ;  but  if  the  answer  is  to  be  less  than  the  third 
term,  make  the  second  less  than  the  first. 

Reduce  the  first  and  second  terms  to  the  lowest  denomination  men- 
tioned in  either,  and  the  third  term  to  the  lowest  denomination  mentioned 
in  it. 

Multiply  the  second  and  third  terms  together,  and  divide  their  product 

*  This  rule  was  formerly  divided  into  the  Rule  of  Three  Direct,  and  the 
Rule  of  Three  Inverse.  The  Rule  of  Three  Direct  included  those  ques- 
tions where  more  required  more  and  less  required  less  ;  thus,  —  If  51b.  of 
coffee  cost  60  cents,  what  would  be  the  value  of  lOlb .?  would  be  a  ques- 
tion in  the  Rule  of  Three  Direct,  because  the  more  coffee  there  was  the 
more  money  it  would  take  to  purchase  it. 

But  if  the  question  were  thus  :  —  If  4  men  can  mow  a  certain  field  in 
12  days,  how  long  would  it  take  8  men  ? —  it  would  be  in  Inverse,  because 
the  more  men  the  less  would  be  the  time  to  perform  the  labor,  that  is, 
•more  would  require  less. 

The  method  for  stating  questions  was  this  :  —  To  make  that  number 
which  is  the  demand  of  the  question  the  third  term,  that  which  is  of  the 
same  name  the  first,  and  that  which  is  of  the  same  name  as  the  answer  re- 
quired, the  second  term  . 

If  the  question  was  direct,  the  second  and  third  terms  must  be  multiplied 
together,  and  their  product  divided  by  the  first ;  but  if  it  was  inverse,  the 
first  and  second  terms  must  be  multiplied  together,  and  their  product  di- 
vided by  the  third. 

18* 


210  PROPORTION.  [SECT.  LII. 

by  tlw  first,  and  the  quotient  is  the  answer,  in  the  same  denomination  to 
which  the  third  is  reduced. 

If  any  thing  remains  after  division,  reduce  it  to  the  next  lower  de- 
nomination, and  divide  as  before. 

If  either  of  tfte  terms  consists  of  fractions,  state  the  question  as  in 
whole  numbers,  and  reduce  the  mixed  numbers  to  improper  fractions, 
compound  fractions  to  simple  ones,  and  invert  the  first  term,  and  then 
multiply  me  three  terms  continually  together,  and  the  product  is  the 
.  answer  to  the  question.  Or  the  fractions  may  be  reduced  to  a  common 
denominator-  and  their  numerators  may  be  used  as  whole  numbers. 
For  tv  hen  fractions  are  reduced  to  a  common  denominator,  their  relative 
value  is  as  their  numerators. 

NOTE  1.  —  In  the  Rule  of  Three,  the  second  term  is  the  quantity 
whose  price  is  wanted  ;  the  third  term  is  the  value  of  the  first  term  ; 
when,  therefore,  the  second  term  is  multiplied  by  the  third,  the  answer 
is  as  much  more  than  it  should  be,  as  the  first  term  is  greater  than  unity  ; 
therefore,  by  dividing  by  the  first  term,  we  have  the  value  of  the  quan- 
tity required".  Or,  multiplying  the  third  by  the  number  of  times  which 
the  second  contains  the  first  will  produce  the  answer. 

NOTE  2.  —  The  pupil  should  perform  every  question  by  analysis,  pre- 
vious to  his  performing  it  by  Proportion. 

EXAMPLES. 

1.  If  a  man  travel  243  miles  in  9  days,  how  far  will  he 
travel  in  24  days  ?  Ans.  648  miles. 

<T    OA       VAI  ^"s  ^e  answer  to  ^6  question  must 

be  in  miles,  we  make  the  third  term 

24  miles  (243)  ;  and  from  the  nature  of 

972  the   question   we   know   that   he  will 

486^  travel  farther  in  24  days  than  in  9  days  ; 

9)5832  we  therefore  place  24,  as  the  larger  of 

Ans7~648  miles.      the  two  remaining  terms,  in  the  second 

place,  and  the  remaining  number,  9 

days,  in  the  first  place. 

CANCELLING. 

24  **"*  =  648  miles.  Ans. 


To  perform  this  question  by  analysis,  we  proceed  thus  :  — 
If  he  travel  243  miles  in  9  days,  he  will  in  one  day  travel  £ 
of  243  miles,  which  is  27  miles  ;  then  if  he  travel  27  miles  in 


SECT.  LII.]  PROPORTION.  211 

one  day,  in  24  days  he  will  travel  24  times  as  far,  which  is 
648  miles,  the  answer,  as  before. 

2.  If  17  yards  of  broadcloth  cost  $102,  what  will  7  yards 
cost  ?  Ans.  $  42. 

yd.     yd.          $ 

17  :  7  :  :  102  As  the  answer  is  to  be  in  dol- 

_7_  lars,   we    make,  the   third   term 

17)714($42  dollars   (102);  and  as  7  yards 

68  WU<1   n°t  cost   so   much    as    17 

oT  yards,  we  make  the  second  term 

34  (7)  less  than  the  first  (17). 

CANCELLING. 

6 
I 

=  $  42  Ans. 


fit 
1 

To  perform  this  question  by  analysis,  we  reason  thus.  If 
17  yards  cost  $102,  one  yard  will  cost  T!T  of  $102,  which  is 
$  6  ;  and  if  one  yard  cost  $  6,  7  yards  will  cost  7  times  as 
much,  which  is  $  42,  the  answer,  as  before. 

3.  If  3  men  drink  a  barrel  of  beer  in  24  days,  how  long 
would  it  last  9  men  ?  Ans.  8  days. 

Men.  m.      da.  AS  the  answer  is  to  be  in  days,  we 

make  24   days  the  third  term,  and   be- 

cause  9  men  will  drink  the  beer  in  less 

9)72  time  than  3  men,  the  second  term  will  be 

8  days.       less  than  the  first. 

To  perform  this  question  analytically,  we  would  say,  that  if 
3  men  drink  a  barrel  of  beer  in  24  days,  it  would  take  one 
man  3  times  24  days,  which  is  72  days  ;  and  if  one  man  drink 
a  barrel  of  beer  in  72  days,  9  men  would  drink  it  in  £  of  72 
days,  which  is  8  days,  as  before. 

4.  If   12   yards   of  cloth   cost   848,  what   will    15  yards 
cost  ?  Ans.  $  60. 

CANCELLING. 


=$60Ans. 


212  PROPORTION.  [SECT.  LII. 

» 

5.  If  17  pounds  of  sugar  cost  $1.19,  what  will  365  pounds 
cost  ?  Ans.  $  25.55. 

6.  If  16  acres  of  land  cost  $720,  what  will   197  acres 
cost  ?  Ans.  $  8865. 

7.  If  $8865  buy   197   acres,  how  many  acres   may   be 
bought  for  $720?  Ans.  16. 

8.  What  will  84hhd.  of  molasses  cost,  if  15hhd.  can  be 
purchased  for  $175.95  ?  Ans.  $  985.32. 

9.  If  $100  gain  $  6  in  12  months,  how  much  would  it  gain 
in  40  months  ?  Ans.  $  20. 

10.  If  a  certain  vessel  has  provisions  sufficient  to  last  a  crew 
of  10  men  45  days,  how  long  would  the  provisions  last  if  the 
vessel  were  to  ship  5  new  hands  ?  Ans.  30  days. 

11.  If  7  and  9  were   12,  what,  on  the  same  supposition, 
would  8  and  4  be  ?  Ans.  9. 

12.  If  9.  men  can  perform  a  certain  piece  of  labor  in  17 
days,  how  long  would  it  take  3  men  to  do  it  ?    Ans.  51  days. 

13.  If  3  men  can  perform  a  piece  of  labor  in  51  days,  how 
many  must  be  added  to  the  number  to  perform  the  labor  in  17 
days  ?  Ans.  6. 

14.  How  much  in  length,  that  is  5£  rods  in  breadth,  is  suffi- 
cient for  an  acre  ?  Ans.  29T*r  rods. 

15.  If  2  barrels  of  flour  cost   $12,  what  will   24    barrels 
cost?  Ans.  $144. 

16.  If  5  quintals  of  fish  cost  $16.25,  what  is  the  value  of  75 
quintals  ?  Ans.  $  243.75. 

17.  If  2  cords  of  wood  cost  $11.50,  what  will    17  cords 
cost  ?  Ans.  .$  97.75. 

18.  If  7cwt.  of  iron  cost  $  56.85,  what  will  49cwt.  cost  ? 

Ans.  $  397.95. 

19.  If  5  acres  of  land  be  valued  at  $  375.75,  what  would  be 
the  value  of  35  acres  ?  Ans.  $  2630.25. 

20.  If  7  pairs  of  shoes  cost  $10.50,  how  many  pairs  will 
$  52.50  buy  ?  Ans.  35. 

21.  If  $4.75  be  paid  for  191b.  of  salmon,  how  many  pounds 
will  $  25.50  buy  ?  Ans.  1021b. 

22.  If  a  man  travels  48  miles  in  6  hours,  how  far  will  he 
travel  in  24  hours  ?  Ans.  192  miles. 

23.  If  8  men  eat  a  barrel  of  flour  in  24  days,  how  long 
would  it  last  3  men  ?  Ans.  64  days. 

24.  If  7  ounces  of  silver  are  sufficient  to  make    17  tea- 
spoons, how  many  may  be  made  from  42  ounces  ? 

Ans.  102.  . 


SECT.  LII.]  PROPORTION.  213 

25.  If  $  100  gain  $  6  in  a  year,   how  much  will    $  850 
gain  ?  Ans.  $  51. 

26.  If  $  100  gain  $  6  in  a  year,  how  much  would  be  suffi- 
cient to  gain  $  32  in  a  year  ?  Ans.  $  533.33£. 

27.  If  20  gallons  of  water  weigh  1671b.,  what  will  180  gal- 
lons weigh  ?  Ans.  15031b. 

28.  If  a  staff  3  feet  long  cast  a  shadow  2  feet,  how  high  is 
that  steeple  whose  shadow  is  75  feet?  Ans.  112£  feet 

29.  If  5-fcwt.  be  carried  36  miles  for  $  4.75,  how  far  might 
it  be  carried  for  $160  ?  Ans.  1212|f  miles. 

30.  If  100  workmen  can  finish  a  piece  of  work  in  12  days, 
how  many  men  are  sufficient  to  finish  the  work  in  8  days  ? 

Ans.  150. 

31.  If  T72-  of  a  yard  cost  fa  of  a  dollar,  what  will  f  of  a 
yard  cost  ?  Ans.  $  0.48. 

A  '?:  •  A        ¥X  f  X  A  =  W*  =  AV  = 

CANCELLING. 
1 

"T  x|x^  =  H  = 

5 

To  perform  this  question  by  analysis,  we  would  proceed 
thus :  —  If  y7^-  of  a  yard  cost  ^  of  a  dollar,  -^  would  cost  j- 
of  $  ^,  which  is  ^ ;  and  if  -fa  cost  rV,  |f  will  cost  12  times 
2^1  which  is  £f  =  f  of  a  dollar.  And  if  one  yard  cost  f  of  a 
dollar,  i  of  a  yard  will  cost  2\  of  a  dollar,  and  f  will  cost  4 
times  as  much,  that  is,  it  will  cost  4  times  ^,  which  is  £f  — 
$  0.48,  as  before. 

32.  If  £  of  a  yard  cost  f  of  a  <£.,  what  will  £  of  a  yard 
cost  ?  Ans.  l£.  Is.  Od. 

33.  If  4£  yards  cost  $  9.75,  what  will  13^-  yards  cost  ? 

Ans.  $  29.25. 

34.  How  much  in  length,  that  is  2J-  inches  wide,  will  make 
a  square  foot  ?  Ans.  57f  inches. 

35.  If  T7^  of  a  ship  cost  51^.,  what  are  -fa  of  her  worth  ? 

Ans.  10o£.  18s.  6fd. 

36.  A  merchant  bought  a  number  of  bales  of  velvet,  each 
containing  129^-  yards,  at  the  rate  of  $  7  for  5  yards,  and  sold 
them  out  again  at  the  rate  of  $1 1  for  7  yards,  and  gained  $  200 
by  the  bargain  ;  how  many  bales  were  there  ?    Ans.  9  bales. 


214  PROPORTION.  [SECT.  LII. 

37.  If  the  moon  moves  13°  NY  35"  in  one  day,  in  what  time 
does  she  perform  one  revolution  ?        Ans.  27du.  7h.  43m. -f- 

38.  If  71b.  of  sugar  cost   £  of  a   dollar,    what   are  121b. 
worth?  Ans.  $1.28£. 

39.  If  $1.75  will  buy  71b.  of  loaf-sugar,  how    much   will 
$213.50  buy  ?  Ans.  7cwt.  2qr.  14lb. 

40.  If  7  ounces  of  gold  are  worth  30c£.,  what  is  the  value  of 
71b.  lloz.  ?  Ans.  407^.  2s.  lOfd. 

41.  My  friend  borrowed  of  me  $  500  for  6  months,  promis- 
ing me  like  favor  ;  soon  after,  I  had  occasion  for  8  600  ;  how 
long  should  I  keep  it  to  receive  full  compensation  for  the  kind- 
ness ?  Ans.  5  months. 

42.  If  the  penny  loaf  weighs  7oz.   when  flour  is  $  8  per 
barrel,  how   much  should  it  weigh  when  flour  is  $  7.50  per 
barrel  ?  Ans.  7^  ounces. 

43.  If  a  regiment  of  soldiers  consisting  of  1000  men  were  to 
be  clothed,  each  suit  containing  3f  yards  of  cloth,  that  is  1£ 
yards  wide,  and  to  be  lined  with  flannel  1£  yards  wide,  how 
many  yards  will  it  take  to  line  the  whole  ?         Ans.  5625yd. 

44.  If  9f  yards  of  broadcloth  cost  $llf,  what  will   163\?y 
ells  English  cost  ?  Ans.  $  24. 

45.  A  merchant  failing  in  trade  owes  in  all  $17280;    his 
effects  are  sold  for  $15120 ;  what  does  he  pay  on  a  dollar,  and 
what  does  A  receive,  to  whom  he  owes  $  5670  ? 

Ans.  He  pays  $  0.874-  on  the  dollar  ;  A  receives  $4961.25. 

46.  Bought  in  London  57  yards  of  broadcloth  for  49  guineas, 
28  shillings  each  ;  what  did  it  cost  per  ell  English  ? 

Ans.  !<£.  10s.  lTyj. 

47.  Bought  a  cask  of  wine,  at  $1.15  per  gallon,  for  $100  ; 
how  much  did  it  contain  ?  Ans.  86gal.  3qt.  14-fpt. 

48.  A  merchant  bought  9  packages  of  cloth  for  $"34560, 
each   package  containing  8  parcels,   each  parcel   12   pieces, 
and  each  piece  20  yards  ;  what  was  the  price  per  yard  ? 

Ans.  $2.00. 

49.  If  75  gallons  of  water  fall  into  a  cistern  containing  500 
gallons,  and  by  a  pipe  in  the  cistern  40  gallons  run  out  in  an 
hour,  in  what  time  will  it  be  filled  ?      Ans.  14h.  17m.  8f  sec. 

50.  How  many  dozen  pairs  of  gloves,  at  $0.56  per  pair, 
may  be  bought  for  $120.96  ?  Ans.  18doz. 

51.  A  certain  cistern  has  three  pipes ;  the  first  will  empty 
it  in  20  minutes,  the  second  in  40  minutes,  and  the  third  in  75 
xninutes  ;  in  what  time  would  they  all  empty  it  ? 

Ans.  llm.  19£§sec. 


SECT.  LII.]  PROPORTION.  215 

52.  A  can  mow  a  certain  field  in  5  days,  and  B  can  mow 
it  in  6  days  ;  in  what  time  would  they  both  mow  it  ? 

Ans.  2T8T  days. 

53.  A  wall,  which  was  to  be  built  32  feet  high,  was  raised  8 
feet  by  6  men  in  12  days ;  how  many  men  must  be  employed 
to  finish  the  wall  in  6  days  ?  Ans.  36  men. 

54.  A  can  build  a  boat  in  20  days,  but  with  the  assistance 
of  C  he  can  do  it  in  12  days ;  in  what  time  would  C  do  it 
himself?  Ans.  30  days. 

55.  In  a  fort  there  are  700  men  provided  with  1840001b.  of 
provisions,  of  which  each  man  consumes  51b.  a  week  ;   how 
long  can  they  subsist  ?  Ans.  52  weeks  4  days. 

56.  If  25  men  have  f  of  a  pound  of  beef  each  three  times  in 
a  week,  how  long  will  31501b.  last  them  ?       Ans.  56  weeks. 

57.  How  many  tiles  8  inches  square  will  lay  a  floor  20  feet 
long,  and  16  feet  wide  ?  Ans.  720. 

58.  How  many  stones  10  inches  long,  9  inches  broad,  and  4 
inches  thick,  would  it  require  to  build  a  wall  80  feet  long,  20 
feet  high,  and  2£  feet  thick  ?  Ans.  17280  stones. 

59.  Bought  threescore  pieces  of  Hollands  for  three  times  as 
many  dollars,  and  sold  them  again  for  four  times  as  many  dol- 
lars ;  but  if  they  had  cost  me  as  much  as  I  sold  them  for,  for 
what  should  I  have  sold  them  to  gain  at  the  same  rate  ? 

Ans.  $  320. 

60.  A  sets  out  on  a  journey,-  and  travels  27  miles  a  day  ;  7 
days  after,  B  sets  out  and  travels  the  same  road  36  miles  a  day ; 
in  how  many  days  will  B  overtake  A?  Ans.  21  days. 

61.  If  I  sell  coffee  at  2s.  3d.  per  Ib.  and  gain  35  per  cent., 
what  did  I  give  per  Ib.  ?  Ans.  Is.  8d. 

62.  A  detachment  of  2000  soldiers  were  supplied  with  bread 
sufficient  to  last  them  12  weeks,  allowing  each  man  14  ounces 
a  day  ;  but  on  examination  find  105  barrels,  containing  2001b. 
each,  wholly  spoiled ;  how  much  a  day  may  each  man  eat,  that 
the  remainder  may  supply  them  12  weeks  ?  Ans.  12oz. 

63.  In  consequence  of  having  a  seventh  part  of  their  bread 
spoiled,  2000  soldiers  were  put  on  an  allowance  of  12  ounces 
of  bread  per  day  for  12  weeks;  what  was  the  whole  weight 
of  their  bread  (good  and  bad),  and  how  much  was  spoiled  ? 

Aris.  The  whole  weight,  1470001b.  ;  spoiled,  210001b. 

64.  Two  thousand  soldiers,  having  lost  105  barrels  of  bread, 
weighing  2001b.  each,  were  obliged  to  subsist  on  12  ounces  a 
day  for  12  weeks ;  but  had  none  been  lost,  they  might  have 
had  14  ounces  a  day  for  the  same  time.     What  was  the  whole 


216  PROPORTION.  [SECT.  LII. 

weight,  including  what  was  lost,  and  how  much  had  they  left 
to  subsist  on  ? 

Ans.   The    whole   weight,    1470001b. ;    left   to   subsist   on, 
1260001b. 

65.  If  2000  soldiers,  after  losing  one  seventh  part  of  their 
bread,  had  each  12  ounces  a  day  for  12  weeks,  what  was  the 
whole  weight  of  their  bread,  including  that  lost,  and  how  much 
might  they  have  had  per  day,  each  man,  if  none  had  been  lost  ? 

Ans.  The  whole  weight  was  1470001b. ;  the  loss,  210001b. ; 
had  none  been  lost,  they  might  have  had  14  ounces  per  day. 

66.  If  .85  of  a  gallon  of  wine  cost  $  2.72,  how  much  will 
.25  of  a  gallon  cost  ?  Ans.  $  0.80. 

67.  If  61.3  pounds  of  tea  cost  $44.9942,  what  is  the  price 
perlb.?  Ans.  $0.73,4. 

68.  What  is  the  value  of  .15  of  a  hogshead  of  lime,  at  $  2.39 
per  hhd.  ?  Ans.  $  0.35,85. 

69.  If  .75  of  a  ton  of  hay  cost  $15,  what  is  it  per  ton  ? 

Ans.  $  20. 

70.  How  many  yards  of  carpeting  that  is  half  a  yard  wide 
will  cover  a  room  that  is  30  feet  long  and  18  feet  wide  ? 

Ans.  120  yards. 

71.  If  a  man  perform  a  journey  in  15  days  when  the  day  is 
12  hours  long,  in  how  many  days  will  he  do  it  when  the  day  is 
but  10  hours  long  ?  Ans.  18  days. 

72.  If  450  men  are  in  a  garrison,  and  their  provisions  will 
last  them  but  5  months,  how  many  must  leave  the  garrison  that 
the  same  provisions  may  be  sufficient  for  those  who  remain  9 
months  ?  Ans.  200  men. 

73.  The  hour  and  minute  hands  of  a  watch  are  together  at 
12  o'clock  ;  when  will  they  next  be  together  ? 

Ans.  Ih.  5m.  27T3j-sec. 

74.  A  and  B  can  perform  a  piece  of  work  in  5T5T  days,  B 
and  C  in  6f  days,  and  A  and  C  in  6  days ;  in  what  time  would 
each  of  them  perform  the  work  alone,  and  how  long  would  it 
take  them  to  do  the  work  together  ? 

Ans.  A  would  do  the  work  in  10  days ;  B,  in  12  days ;  C,  in 
15  days ;  A,  B,  and  C,  together,  in  4  days. 

75.  A,  B,  and  C  can  perform  a  piece  of  work  in  4  days,  B 
can  do  it  in  12  days,  C  can  do  it  in  15  days ;  in  what  time 
would  A  and  B  perform  the  labor  ?  Ans.  5yV  days. 

76.  How  many  bricks  8  inches  long,  4  inches  wide,  and  2 
inches  thick,  will  it  require  to  build  the  walls  of  a  house  which 
is  46  feet  long,  28  feet  wide,  and  25  feet  high,  and  the  walls  to 
be  18  inches  thick  ?  Ans.  143,775 


SECT.  LIII.]  COMPOUND  PROPORTION.  217 

77.  Lent  a  friend  $  200  for  12  months,  on  condition  of  his 
returning  the  favor;  how  long  ought  he  to  lend  me  $150  to 
requite  my  kindness  ?  Ans.  16  months. 

78.  If  5  oxen  or  7  cows  eat  3^T  tons  of  hay  in  87  days,  in 
what  time  will  2  oxen  and  3  cows  eat  the  same  quantity  of 
hay  ?  Ans.  105  days. 

79.  If  360  men  be  placed  in  a  garrison,  and  have  provisions 
for  6  months,  how  many  men  must  be  sent  away  at  the  end 
of  4  months  that  the  remaining  provision  may  last  them  8 
months  longer  ?  Ans.  270  men. 

80.  My  tailor  informs  me  it  will  take  10£  square  yards  of 
cloth  to  make  me  a  full  suit  of  clothes.     The  cloth  I  am  about 
to  purchase  is  If  yards  wide,  and  on  sponging  it  will  shrink  5 
per  cent,  in  width  and  length.     How  many  yards  of  the  above 
cloth  must  I  purchase  for  my  "  new  suit "  ?        Ans.,  6T§§ 7yd. 


SECTION  LIII. 
COMPOUND   PROPORTION, 

OR 

DOUBLE  RULE  OF  THREE. 

COMPOUND  PROPORTION  is  the  method  of  performing  such 
operations  in  Proportion  as  require  two  or  more  statements. 

EXAMPLES. 

1.  If  a  man  travel  117  miles  in  30  days,  employing  only  9 
hours  a  day,  how  far  would  he  go  in  20  days,  travelling  12 
hours  a  day  ? 

The  distance  to  be  travelled  depends  on  two  circumstances, 
—  the  number  of  days  the  man  travels,  and  the  number  of 
hours  he  travels  in  each  day. 

We  will  first  suppose  the  hours  to  be  the  same  in  each  case ; 
the  question  will  then  be,  —  If  a  man  travel  1 17  miles  in  30 
days,  how  far  will  he  travel  in  20  days  ? 

This  will  lead  to  the  following  proportion. 

30  days  :  20  :  :  117  miles  :  ^p?  =  73  miles. 
19 


218  COMPOUND  PROPORTION.  [SECT.  MH. 

That  is,  if  we  multiply  1 17  by  20,  and  divide  the  product  by 
30,  we  obtain  the  number  of  miles  he  will  travel  in  20  days, 
which  is  78. 

Now,  if  we  take  into  consideration  the  number  of  hours,  we 
must  say,  —  If  a  man,  travelling  9  hours  a  day  for  a  certain 
number  of  days,  has  travelled  78  miles,  how  far  will  he  go  in 
the  same  time,  if  he  travel  12  hours  a  day  ?  This  will  furnish 
the  following  proportion. 

9  hours  :  12  hours  :  :  78  miles  :  l^^  =  104  miles,  the 
answer  to  the  question. 

By  this  mode  of  resolving  the  question,  we  see  that  117  miles 
have,  to  the  answer  104  miles,  the  proportion  that  30  days 
have  to  20  days,  and  that  9  hours  have  to  12  hours.  Stating 
this  in  Compound  Proportion,  we  have 

39  •  12  I  :  :  117  :  104  miles' the  answer- 

Thus  it  appears  that  if  1 17  be  multiplied  by  both  20  and  12, 
and  the  product  be  divided  by  30  times  9,  the  quotient  will  be 
104  miles ;  or  if  we  multiply  1 17  by  20,  and  divide  the  product 
by  30,  and  then  multiply  this  quotient  by  12  and  divide  by  9, 
it  will  produce  the  same  answer  as  before. 

This  question  may  be  performed  by  analysis  thus  :  —  If  he 
travel  117  miles  in  30  days,  in  one  day  he  will  travel  -fa  of 
1 17  miles,  which  is  ty?-  miles  ;  and,  travelling  9  hours  a  day, 
he  will  in  one  hour  travel  £  of  J^y-  miles,  which  is  £$  miles  ; 
and  in  a  day  of  12  hours  he  will  travel  12  times  £§  miles, 
which  is  -Vtf6-  miles  ;  and  in  20  days  he  will  travel  20  times 
Jg^6-  miles,  which  is  104  miles,  the  answer,  as  before. 

The  answer  to  the  above  question  might  have  been  obtained 
by  dividing  the  third  term  by  the  product  of  the  two  ratios 
which  the  first  two  terms  have  to  the  second  terms ;  that  is,  by 
the  ratio  of  30  to  20,  which  is  J g  —  £ ;  and  of  9  to  12,  which 
wA  =  J.  Thus, 

1 17  ^_  £  x  £  =  1 17  -4- 1  =  H1  X  f  =  *f A  —  104  Ans. 

2.  If  6  men  in  16  days  of  9  hours  each  build  a  wall  20  feet 
long,  6  feet  high,  and  4  feet  thick,  in  how  many  days  of  8 
hours  each  will  24  men  build  a  wall  200  feet  long,  8  feet  high, 
and  6  feet  thick  ? 

In  stating  this  question,  there  are  several  circumstances  to 
be  taken  into  consideration  ;  the  number  of  men  employed, 


SECT.  LIII.]  COMPOUND  PROPORTION.  219 

the  length  of  the  days,  length  of  the  wall,  and  its  height  and 
breadth. 

As  the  answer  to  the  question  is  to  be  in  days,  we  make  the 
days  the  third  term. 

Were  all  the  circumstances  of  the  question  alike,  except  the 
number  of  men  and  the  number  of  days,  the  question  would 
consist  in  finding  in  how  many  days  24  men  would  perform 
the  same  labor  that  6  men  had  done  in  16  days  ;  that  is,  if  6 
men  had  built  a  certain  wall  in  16  days,  how  many  days  would 
it  take  24  men  to  perform  the  same  labor  ?  This  would  furnish 
the  following  proportion. 

24  men  :  6  men  :  :  16  days  :  6-^-6  =  4  days. 

Or,  if  this  were  the  question,  —  If  a  certain  number  of  men, 
by  laboring  9  hours  a  day,  perform  a  piece  of  work  in  16  days, 
how  many  days  would  it  take  the  same  men  to  do  the  labor  by 
working  8  hours  a  day  ?  —  the  following  would  be  the  propor- 
tion. 


8  hours  :  9  hours  :  :  16  days  :  -       =  18  days. 

Or,  if  this  were  the  question,  —  If  a  certain  number  of  men 
build  a  wall  20  feet  long  in  16  days,  how  long  would  it  take  the 
same  men  to  build  a  wall  200  feet  long  ?  —  the  following  would 
be  the  statement. 


20  feet  :  200  feet  :  :  16  days  :  -  160  days. 

Or,  if  only  the  days  and  height  of  the  wall  were  considered, 
this  would  be  the  statement. 

6  feet  :  8  feet  :  :  16  days  :  ^-6  —  21£  days. 

Lastly,  were  we  to  consider  only  the  days  and  the  thickness 
of  the  wall,  it  would  furnish  the  following  statement. 

4  feet  :  6  feet  :  :  16  days  :  6-^  —  24  days. 

We  see,  by  this  mode  of  resolving  the  question,  that  16  days 
must  have  to  the  true  answer  the  ratio  compounded  of  the 
ratios 

That  24  men  have  to  6  men  ; 
That  8  hours  have  to  9  hours  ; 
That  20  feet  have  to  200  feet  ; 
That  6  feet  have  to  8  feet  ;  and 
That  4  feet  have  to  6  feet. 
Stating  the  above  in  Compound  Proportion,  we  have 


220 


COMPOUND  PROPORTION. 


[SECT.  LIII. 


24  men 

8  hours 
20  feet 
6  feet 
4  feet 


6  men 
9  hours 
200  feet 
8  feet 
6  feet 


: :  16  days  :  90  days  =  Answer. 


The  continued  product  of  all  the  second  terms  by  the  third 
term,  and  this  divided  by  the  continued  product  of  the  first 
terms,  will  produce  the  answer. 


Thus 


6X9X200X8X6X16 
21X8X-20X6X4 


=  90  days,  Answer. 


OPERATION    BY    CANCELLING. 
10  $ 


0  X  9  X  #00  X  0  X  6  X 


3.  If  5  compositors  in  16  days,  11  hours  long,  can  compose 
25  sheets  of  24  pages  in  each  sheet,  and  44  lines  in  a  page, 
and  40  letters  in  a  line,  in  how  many  days  10  hours  long  may 
9  compositors  compose  a  volume,  to  be  printed  on  the  same  let- 
ter, consisting  of  36  sheets,  16  pages  to  a  sheet,  50  lines  to  a 
page,  and  45  letters  in  a  line  ?  Ans.  12  days. 


STATEMENT. 

9  comp.  :  5  comp. 
10  hours  :  11  hours 
25  sheets  :  36  sheets 
24  pages  :  16  pages 
44  lines  :  50  lines 
40  letters :  45  letters  ' 


16  days  :  12  days,  Ans. 


OPERATION    BY    CANCELLING, 


$        4 


'  Ans' 


RULE.  —  Make  that  number  which  is  of  the  same  kind  as  the  answer 
required  the  third  term  ;  and  of  the  remaining  numbers,  take  any  two 
that  are  of  the  same  kind,  and  consider  whether  an  ansioer  depending 
upon  these  alone  would  be  greater  or  less  than  the  third  term,  and  place 
them  as  directed  in  Simple  Proportion.  Then  take  any  other  two,  and 
consider  whether  an  answer  depending  only  upon  them  would  be  greater 
or  less  than  the  third  term,  and  arrange  them  accordingly  ;  and  so  on,. 


SECT.  LIV.]  CHAIN   RULE. 

until  all  are  used.  Multiply  the  continued  pro 
by  the  third,  and  divide  by  the  continued  prod 
produce  the  answer. 

NOTE.  — All  the  following  questions  are  to  be  performed  not  only  by 
the  Rule,  but  by  analysis.    The  pupil  should  also  apply  the  cancelling  rule. 

4.  If  $100  gain  $6  in  one  year,  how  much  would  $500 
gain  in  four  months  ?  Ans.  $10. 

5.  If  $100  gain  $  6  in  one  year,  what  must  be  the  sum  to 
gain  $10  in  4  months  ?  Ans.  $  500. 

6.  How  long  will  it  take  $  500  to  gain  $10,  if  $100  gain 
$  6  in  one  year  ?  Ans.  4  months. 

7.  If  $  500  gain  $  10  in  4  months,  what  is  the  rate  per 
cent.  ?  Ans.  6  per  cent. 

8.  If  8  men  spend  $  32  in  13  weeks,  what  will  24  men 
spend  in  52  weeks  ?  Ans.  $  384. 

9.  If  12  men  can  build  a  wall  30  feet  long,  6  feet  high,  and 
3  feet  thick,  in  15  days,  when  the  days  are  12  hours  long,  in 
what  time  will  60  men  build  a  wall  300  feet  long,  8  feet  high, 
and  6  feet  thick,  when  they  work  only  8  hours  a  day  ? 

Ans.  120  days. 

10.  If  16  horses  consume  84  bushels  of  grain  in  24  days, 
how  many  bushels  will  suffice  32  horses  48  days  ? 

Ans.  336  bushels. 

11.  If  the  carriage  of  5cwt.  3qr.  150  miles  cost  $24.58, 
what  must  be  paid  for  the  carriage  of  7cwt.  2qr.  251b.  64  miles, 
at  the  same  rate  ?  Ans.  $14.08,6. 

12.  If  7£oz.  of  bread  be  bought  for  4f  d.  when  corn  is  4s.  2d. 
per  bushel,  what  weight  of  it  may  be  bought  for  Is.  2d.  when 
the  price  per  bushel  is  5s.  6d.  ?  Ans.  16££f  oz. 

13.  If  496  men,  in  5£  days  of  1 1  hours  each,  dig  a  trench 
of  7  degrees  of  hardness   465   feet  long,  3|  wide,  2£  deep,  in 
how  many  days  of  9  hours  long  will  24  men  dig  a  trench  of  4 
degrees  of  hardness  337£  feet  long,  5f  wide,  and  3£  deep  ? 

Ans.  132  days. 


SECTION  LIV. 
CHAIN   RULE. 

THE  CHAIN  RTTLE  consists  in  joining  many  proportions  to- 
gether, and  by  the  relation  which  the  several  antecedents  have 
19* 


222  CHAIN   RULE.  [SECT.  LIV. 

to  their  consequents  the  proportion  between  the  first  antecedent 
and  the  last  consequent  is  discovered. 

This  rule  may  often  be  abridged  by  cancelling  equal  quanti- 
ties on  both  sides,  and  abbreviating  commensurables. 

NOTE.  —  The  first  numbers  in  each  part  of  the  question  are  called  an- 
tecedents, and  the  following  consequents. 

1.  If  201b.  at  Boston  make  231b.  at  Antwerp,  and  1551b.  at 
Antwerp  make  1801b.  at  Leghorn,  how  many  pounds  at  Boston 
are  equal  to  1441b.  at  Leghorn  ? 

OPERATION.  It  will  be  perceived 

201b.  of  Boston     =    23  of  Antwerp,  in  the  operation  that 

1551b.  of  Antwerp  =  180  of  Leghorn,  the  continued  product 

1441b  of  Leghorn.  of  the  antecedents  is 

20x155x144       446400        nyy,91k   A  divided    by   the   con- 

23X160          LIT!^  :      lOTiJlb.  Ans.       tinued  product  of  the 

consequents. 

CANCELLING. 

16 
20  X  155  X 


33  X 

0 

RULE.  —  Write  tJie  numbers  alternately  ,  that  is,  the  antecedents  at  tfo 
left  hand,  and  the  consequents  at  the  right  ;  and,  if  the  last  number  stands 
at  the  left  hand,  multiply  the  numbers  of  the  left-hand  column  continually 
together  for  a  dividend,  and  those  at  tlie  right  for  a  divisor  ;  but,  if  tJie 
last  nunibcr  stands  at  the  right  hand,  multiply  the  numbers  of  the  right- 
hand  column  continually  together  for  a  dividend,  and  those  at  the  left  for 
a  divisor  ;  and  the  quotient  will  be  the  answer. 

NOTE.  —  The  demonstration  for  this  rule  is  the  same  as  for  Compound 
Proportion. 

2.  If  121b.  at  Boston  make  lOlb.  at  Amsterdam,  and  lOlb.  at 
Amsterdam  make  121b.  at  Paris,  how  many  pounds  at  Boston 
are  equal  to  801b.  at  Paris  ?  Ans.  801b. 

CANCELLING. 

M  X  W  X  80 


3.  If  251b.  at  Boston  are  equal  to  221b.  at  Nuremburg,  and 
881b.  at  Nuremburg  are  equal  to  921b.  at  Hamburg,  and  461b. 
at  Hamburg  are  equal  to  491b.  at  Lyons,  how  many  pounds  at 
Boston  are  equal  to  981b.  at  Lyons  ?  Ans.  lOOlb. 


SECT.  LV.]  PARTNERSHIP.  223 

CANCELLING. 

4  £ 

25  X  W  X  ^0  X  00 


NOTE.  —  The  pupil  may  cancel  all  the  following  questions  in  a  similar 
manner. 

4.  If-  24  shillings  in  Massachusetts  are  equal  to  32  shillings 
in  New  York,  and  if  48  shillings  in  New  York  are  equal  to  45 
shillings  in  Pennsylvania,  and  if  15  shillings  in  Pennsylvania 
are  equal  to  10  shillings  in  Canada,  how  many  shillings  in  Can- 
ada are  equal  to  100  shillings  in  Massachusetts  ? 

Ans.  83£  shillings. 

5.  If  17  men  can  do  as  much  work  as  25  women,  and  5 
women  do  as  much  as  7  boys,  how  many  men  would  it  take  to 
do  the  work  of  75  boys  ?  Ans.  36f  men. 

6.  If  10  barrels  of  apples  will  pay  for  5  cords  of  wood,  and 
20  cords  of  wood  for  4  tons  of  hay,  how  many  barrels  of  apples 
will  it  take  to  purchase  50  tons  of  hay  ?  Ans.  500bbl. 

7.  If  100  acres  in  Bradford  be  worth  120  in  Haverhill,  and 
50  in  Haverhili  be  worth  65  in  Methuen,  how  many  acres  in 
Bradford  are  equal  to  150  in  Methuen  ?        Ans.  96T27  acres. 

8.  If  lOlb.  of  cheese  are  equal  in  value  to  71b.  of  butter, 
and  lllb.  of  butter  to  2  bushels  of  corn,  and  11  bushels  of  corn 
to  8  bushels  of  rye,  and  4  bushels  of  rye  to  one  cord  of  wood, 
how  many  pounds  of  cheese  are  equal  in  value  to  10  cords  of 
wood  ?  Ans.  432  jib. 


SECTION  LV. 
PARTNERSHIP,  OR  COMPANY  BUSINESS. 

PARTNERSHIP  is  the  association  of  two  or  more  persons  in. 
business,  with  an  agreement  to  share  the  profits  and  losses  in 
proportion  to  the  amount  of  the  capital  stock  contributed  by 
each. 

EXAMPLES. 

1.  Three  men,  A,  B,  and  C,  enter  into  partnership  for  two 
years,  with  a  capital  of  $1080.  A  puts  in  $240,  B  $  360,  and 


224  PARTNERSHIP.  [SECT.  LV. 

C  $  480.     They  gain  $  54.     What  is  each  man's  share  of  the 
gain  ? 

As  the  whole  stock  in  trade  is  $  1080,  of  which  $  240  be- 
longs to  A,  A's  share  of  the  stock,  therefore,  will  be  -fiff^  —  f , 
and  as  each  man's  gain  is  in  proportion  to  his  stock,  f  of  $  54, 
which  is  $12,  is  A's  gain.  B's  stock  in  trade  is  $  360  ;  there- 
fore fVW  =  i  of  $  54,  which  is  $18,  is  B's  gain.  C's  stock  is 
$  480  ;  therefore  his  part  of  the  whole  stock  is  -fgfo  =  f  ;  con- 
sequently C's  share  of  the  gain  is  £  of  $  54,  which  is  $  24. 
Hence,  to  find  any  man's  gain  or  loss  in  trade  we  have  the  fol- 
lowing 

RULE.  —  Multiply  the  whole  gain  or  loss  by  each  man's  fractional 
fart  of  the  stock. 

NOTE.  —  The  pupil  who  may  be  desirous  of  performing  the  questions 
of  this  rule  in  the  "  old  icay  "  will  adopt  the  following 

RULE.  —  As  the  whole  stock  is  to  the  whole  gain  or  loss,  so  is  each 
man's  particular  stock  to  his  particular  share  of  the  gain  or  loss. 

The  following  is  the  statement  of  the  first  question,  with  the 
answers  and  proof. 

As  the  stock  $1080  :  $  54  ::  $  240  :  $12  A's  gain. 

$1080  :  $  54  ::  $  360  :  $18  B's  gain. 

$1080  :  $  54  : :  $480  :  $24  C's  gain. 

Proof,  $12  +  $18  +  $  24  =  $  54  whole  gain. 

2.  A,  B,  and  C  enter  into  partnership,  with  a  capital  of 
$1100,  of  which  A  put  in  $  250,  B  put  in  $  300,  and  C  $  550 ; 
they  lost  by  trading  5  per  cent,  on  their  capital.     What  was 
each  man's  share  of  the  loss  ?  A's  loss,  $  12.50. 

B's  loss,  $  15.00. 
C's  loss,  $27.50. 

3.  Two  merchants,  C  and  D,  engaged  in  trade  ;  C  put  in 
$6780,  and  D  put  in  $12,000;  they  gain  $1000.     What  is 
each  man's  share  ?  C's  share,  $361.02,2£||. 

D's  share,  $  638.97,7£f  f . 

4.  M,  P,  and  Q  trade  in  company,  with  a  capital  of  $10,000  ; 
M  put  in  8  3000,  P  $  2000,  and  Q  $  5000  ;  they  gain  $  500. 
What  is  each  man's  share  of  the  gain  ?          M's  gain,  $150. 

P's  gain,  $100. 
Q's  gain,  $250. 

5.  A,  B,  and  C  enter  into  partnership  ;    A  put  in  $  500,  B 
$  350,  and  C  put  in  320  yards  of  broadcloth  ;  they  gain  $  332.50, 


SECT.  LVI.]  PARTNERSHIP  ON   TIME.  225 

of  which  C's  share  is  $120.     What  were  A's  and  B's  shares  of 
the  gain,  and  what  was  the  value  of  C's  cloth  per  yard  ? 

A's  gain,  $125.00. 

B's  gain,  $87.50. 

C's  cloth  per  yd.  $1.50. 

6.  A,  B,  and  C  trade  in  company*;  A  put  in  $  5000,  B  put  in 
$  6500,  and  C  put  in  $  7500  ;  they  gain  40  per  cent,  on  their 
capital,  but  receive  the  whole  amount  of  their  gains  in  bills, 
for  which  they  are  obliged  to  allow  a  discount  of  10  per  pent. 
How  much  was  each  man's  net  gain  ? 

A's  gain,  $1800. 
B's  gain,  $2340. 
C's  gain,  $  2700. 

7.  A  merchant,  failing  in  trade,  owes  A  $  600,  B  $  760,  C 
$  840,  and  D  $  800.     His  effects  are  sold  for  $  2275.     What 
will  each  receive  of  the  dividend  ?  A,  $  455.00. 

B,  $  576.33£. 

C,  $  637.00. 

D,  $  606.66f . 

8.  A  bankrupt  owes  $  5000.     His  effects,  sold  at  auction, 
amount  to  $  4000.     What  will  his  creditors  receive  on  a  dollar  ? 

Ans.  $  0.80. 

9.  A  merchant,  having  sustained  many  losses,  is  obliged  to 
become  a  bankrupt.     His  effects  are  valued  at  $1728,  with 
which  he  can  pay  only  fifteen  cents  on  the  dollar.     What  did 
he  owe  ?  Ans.  $11,520. 


SECTION  LVI. 
PARTNERSHIP   ON   TIME. 

WHEN  merchants  in  partnership  employ  their  stock  for  une- 
qual times,  it  is  called  Partnership  on  Time. 

EXAMPLES. 

1.  Two  men,  A  and  B,  trade  in  company.  A  puts  in  $  420 
for  5  months,  and  B  puts  in  $  350  for  8  months  ;  they  gain  $  84. 
What  is  each  man's  share  of  the  gain  ? 

METHOD    OF    OPERATION    BY   ANALYSIS. 

$  420  for  5  months  is  the  same  as  5  x  $  420  =  $  2100  for  1 


226  PARTNERSHIP  ON  TIME.  [SECT.  LVI. 

month  ;  and  $  350  for  8  months  is  the  same  as  8  X  $  350  = 
$  2f?00  for  1  month.  The  question,  therefore,  is  the  same  as 
if  A  had  put  in  $  2100  and  B  $  2800  for  1  month  each.  The 
whole  stock  would  then  be  $  4900.  A's  share  of  the  gains, 
therefore,  will  be  £  £$£  ==  f  of  $  84  =  $  36.  And  B's  share 
will  be  £-  =  of  $  84  =  $  48. 


RULE.  —  Multiply  each  man's  stock  by  tJie  time  it  continued  in  trade, 
and  consider  cacti  product  a  numerator  to  be  written  over  the  sum  of  all 
the  numerators,  as  a  common  denominator  ;  then  multiply  the  whole  gain 
or  loss  by  each  fraction,  and  the  several  products  will  be  the  gain  or  loss 
of  each  man, 

NOTE.  —  It  might  be  well  for  the  student  to  be  acquainted  with  the 
u  old  way  "  of  performing  these  questions.  The  following  is  the 

RULE.  —  Multiply  each  man's  stock  by  the  time  it  was  continued  in 
trade  ;  then,  as  the  sum  of  all  the  products  is  to  the  whole  gain  or  loss, 
so  is  each  man's  "particular  product  to  his  share  of  the  gain  or  loss. 

The  first  question  would  be  performed  thus  : 
$420  X  5  =  2100     4900  :  84  ::  2100  :  836  A's  gain. 
$  350  X  8  =  2800     4900  :  84  :  :  2800  :  $  48  B's  gain. 
4900 

Proof,  $  36  +  $  48  =  $  84  =  A  and  B's  gain. 

2.  A  commenced  business,  January  1,  with  a  capital  of  $  3200  ; 
May  1,  he  took  B  into  partnership,  with  a  capital  of  $4200  ; 
and  at  the  end  of  the  year  they  had  gained  $  240.  What  was 
each  man's  share  of  the  gain  ?  $  128,  A's  gain. 

$112,  B's  gain. 

3.  A,  B,  and  C  traded  in  company.     A  put  in  $  300  for  5 
months,  B  put  in  $  400  for  8  months,  and  C  put  in  $  500  for  3 
months  ;  they  gain  $100.     What  is  the  gain  of  each  ? 

$24.19^|,  A's  gain. 
$  51.61&,  B's  gain. 
$24.19£|,  C'sgain. 

4.  Three  men  hire  a  pasture  in  common,  for  which  they  are 
to  pay  $  26.40.     A  put  in  24  oxen  for  8  weeks,  B  put  in  18 
oxen  for  12  weeks,  and  C  put  in  12  oxen  for  10  weeks.     What 
ought  each  to  pay  ?  A  should  pay  $  9.60. 

B  should  pay  $10.80. 
C  should  pay  $  6.00. 

5.  Two  men  in  Boston  hire  a  carriage  for  $  25,  to  go  to  Con- 
cord, N.  EL,  the  distance  being  72  miles,  with  the  privilege  of 
taking  in  three  more  persons.     Having  gone  20  miles,  they 


SECT.  LVII.]  GENERAL   AVERAGE.  227 

take  in  A ;  at  Concord,  they  take  in  B  ;  and  when  within  30 
miles  of  the  city,  they  take  in  C.  How  much  shall  each  man 
pay  ?  First  man  pays  $  7.60,9|£f . 

Second  man  pays  $  7.60,9|&f . 

A  pays  $  5.87,3TVg-. 

B  pays  $  2.86,4/2-. 

C  pays  $  1.04,1TV 

$  25.007T" 

6.  Three  men  engage  in  partnership  for  20  months.     A  at 
first  put  into  the  firm  $  4000,  and  at  the  end  of  four  months 
he  put  in  $  500  more,  but  at  the  end  of  16  months  he  took  out 
$1000  ;    B  at  first  put  in  $  3000,  but  at  the  end  of  10  months 
he   took  out  $  1 500,  and  at  the  end  of  14  months  he  put  in 
$  3000  ;  C  at  first  put  in  $  2000,  and  at  the  end  of  6  months 
he  put  in  $  2000  more,  and  at  the  end  of  14  months  he  put  in 
$2000  more,  but  at  the  end  of  16  months  he  took  out  $1500  ; 
they  had  gained  by  trade  $  4420.     What  is  each  man's  share 
of  the  gain  ?  A's  gain,  $1680. 

B's  gain,  $1260. 
C's  gain,  $1480. 

7.  John  Jones,  Samuel  Eaton,  and  Joseph  Brown  formed  a 
partnership,  under  the  firm  of  Jones,  Eaton,  &  Co.,  with  a  cap- 
ital of  $10,000 ;  of  which  Jones  put  in  $4000,  Eaton  put  in 
$  3500,  and  Brown  put  in  $  2500  ;  but  at  the  end  of  6  months 
Jones  withdrew  $  2000  of  his  stock,  and  at  the  end  of  8  months 
Eaton  withdrew  $1500  from  the  firm;   but  at  the  end  of  10 
months  Brown  added  $  2000  to  his  stock.     At  the  end  of  2 
years  they  found  their  gains  to  be  $1041.80.     What  was  the 
share  of  each  man  ?  Jones's    gain,  $  300.51^f. 

Eaton's  gain,  $300.51|§. 
•  Brown's  gain,  $  440.76T27. 


SECTION  LVII. 
GENERAL   AVERAGE. 

GENERAL  AVERAGE  is  in  mercantile  law  whatever  damage 
or  loss  is  incurred  by  any  part  of  a  ship  and  cargo  for  the  pres- 
ervation of  the  rest. 

When  real  damage  occurs,  the  several  persons  interested  in 
the  ship,  freight,  and  cargo  contribute  their  respective  propor- 
tions to  indemnify  the  owners  of  the  part  in  question  against  the 


228  GENERAL  AVERAGE.  [SECT.  LVII. 

damage  or  necessary  expense  which  has  been  incurred  for  the 
good  of  all. 

No  general  average  takes  place  unless  the  danger  was  immi- 
nent, and  absolutely  necessary  for  the  safety  of  the  ship. 

'  Particular  Average  signifies  a  partial  loss  of  the  ship  and 
cargo,  arising  from  accidents  at  sea,  and  the  loss  must  be  borne 
by  the  owners  of  the  property,  who  have  sustained  damage. 
Underwriters  must  pay  such  proportion  of  the  prime  cost  as 
corresponds  with  the  proportion  of  diminution  in  value  occa- 
sioned by  the  damage. 

Goods,  whether  lost,  injured,  or  saved,  are  to  be  valued  at  the 
price  they  would  have  brought  in  ready  money  on  the  ship's  ar- 
riving at  her  destined  port.  The  value  of  the  vessel  is  also  es- 
timated in  the  same  manner. 

It  is  a  general  custom  to  allow  in  the  general  average  two 
thirds  of  the  expense  incurred  in  procuring  new  masts,  cables, 
and  other  furniture  of  the  ship,  the  new  materials  being  much 
better  than  the  old.  A  deduction  in  the  general  average  of  from 
one  third  to  one  half  is  made  from  the  seamen's  wages  while 
the  vessel  is  detained  in  port.  Goods  are  valued  at  the  invoice 
price  when  the  average  claim  is  settled  at  the  port  of  lading. 

To  find  the  General  Average. 

RULE.  —  As  the  value  of  all  tJie  articles  subject  to  contribution  is  to 
the  whole  loss,  so  is  each  person's  share  of  that  value  to  his  average  pro- 
portion of  tfe  loss ;  or  so  is  $  100  to  the  loss  per  cent. 

The  ship  General  Taylor,  in  her  voyage  from  Boston  to  Mo- 
bile, on  the  night  of  January  10,  1847,  was  wrecked  on  Nan- 
tucket  Shoals  ;  in  consequence  of  which  the  captain  was  obliged 
to  cut  away  her  masts,  and  heave  overboard  some  of  the  furni- 
ture of  the  ship  and  part  of  the  cargo.  The  vessel  was  finally 
towed  into  New  York  by  the  steamboat  Ohio.  A  statement  of 
the  loss  and  expenses  is  as  follows  :  — 

Amount  of  tJte  Loss. 

Goods  of  R.  S.  Davis  cast  overboard,          .         .     $1728 

Damage  of  J.  Smith's  goods,  ....     772 

"       "     C.  Dana's  goods,        ....         866 

Amount  of  the  freight  of  goods  thrown  overboard,        334 

"       "     two  thirds  of  the  expense  in  procuring 

new  masts,  cable,  anchors,  &c., 

Pilotage  and  port  duties,      ....  400 

Miscellaneous  expenses,          .....       25 

$5000 


SECT.  LVIII.]  PROFIT   AND  LOSS.  229 

Contributary  Articles. 

Value  of  the  ship, .     $  9272 

Do.  of  R.  S.  Davis's  goods  thrown  overboard,  .  .  1728 
Do.  of  J.  Smith's  goods  at  Mobile,  deducting  freight,  2000 
Do.  of  C.  Dana's  goods  do.  do.  7000 

Do.   of  two  thirds  the  ship's  freight,   .         .         .         .     5000 

$  25,000 

A  Statement  of  what  each  Contributary  Article  pays  in  the  General 
Average. 

The  ship,  $9272  X  -20  pays  $1854.40 

Ship's  freight,        5000  X  .20     "         1000.00 
Davis's  goods,       1728  X  -20     "  345.60 

Smith's  goods,      2000  X  .20      "          400.00 
Dana's  goods,       7000  X  .20     "         1400.00 
8  25,000  $  5000.00 

A  Statement  of  what  each  Party  receives. 
Davis  receives  $1728  X  .80  =  $1382.40 
Smith      do.          772  X  .80  =      617.60  and  his  goods. 
Dana       do.          866  X  .80  =      692.80     do.         do. 
Owners  receive  2307.20 

$5000.00 


SECTION  LVIII. 
PROFIT    AND   LOSS. 

BY  this  rule  merchants  estimate  their  profit  and  loss  in  buy- 
ing and  selling  goods. 

The  questions  are  to  be  performed  by  Proportion  and  the  oth- 
er preceding  rules.  The  pupil  should  give  an  analysis  of  each 
question. 

One  of  four  things  is  generally  required  by  this  rule  :  — 

1.  To  know  what  per  cent,  is  gained  or  lost,  either  in  pur- 
chasing or  selling  goods. 

2.  To  ascertain  at  what  price  to  sell  an  article  to  gain  or  lose 
a  certain  per  cent. 

3.  To  ascertain  the  price  of  an  article,  when  we  know  what 
per  cent,  is  gained  or  lost. 

4.  If  goods  be  sold  at  a  certain  price,  and  there  be  gained 

20 


230  PROFIT  AND  LOSS.  [SECT.  LVHI. 

or  lost  a  certain  per  cent.,  to  ascertain  what  would  be  gained  or 
lost  at  some  other  price. 

The  eight  following  examples  will  illustrate  the  above  prob- 
lems. 

EXAMPLES. 

1.  If  I  buy  cloth  at  $  4  per  yard,  and  sell  it  at  $  5  per  yard, 
what  per  cent,  do  I  gain  ?  Ans.  25  per  cent. 

OPERATION    BY    PROPORTION. 

$  5  _  $  4  —  $1 ;  $  4  :  $1  : :  $100  :  $  25,  that  is,  25  per  cent. 

By  analysis.  If  $4  gain  $1,  it  is  evident  that  $1  will  gain 
|  of  $1  =  $0.25;  and  $100  will  gain  100  times  $0.25  = 
$25,  that  is,  25  per  cent.,  answer  as  before. 

2.  When  cloth  is  purchased  at  $  5  per  yard,  and  sold  at  $  4 
per  yard,  what  per  cent,  is  lost  ?  Ans.  20  per  cent. 

OPERATION    BY    PROPORTION. 

$  5  —  $  4  =  $1.     $5  :  $1  : :  $100  :  $  20,  that  is,  20  per  cent. 

By  analysis.  If  $  1  be  lost  on  $  5,  it  is  certain  that  on  $  1 
there  will  be  lost  |  of  $1  =  $  0.20  ;  and  if  on  $1  there  be  lost 
$0.20,  on  $100  there  will  be  lost  100  times  $  0.20  =  $20,  that 
is,  20  per  cent.,  answer  as  before  ;  therefore,  when  we  wish  to 
know  what  per  cent,  is  gained  or  lost,  either  in  purchasing  or 
disposing  of  goods,  we  adopt  the  following 

RULE.  —  As  the  price  of  the  goods  is  to  the  gain  or  /oss,  so  is  $  100 
to  the  per  cent,  gained  or  lost. 

3.  If  I  buy  cloth  at  $  4  per  yard,  for  how  much  must  I  sell  it 
to  gain  25  per  cent.  ?  Ans.  $  5. 

OPERATION    BY    PROPORTION. 

$100 +  $25  =  $125;  $100:  $125::  $4:  $5  Ans. 

By  analysis.  If  for  $100  I  receive  $125,  it  is  evident  that 
for  $1  I  shall  receive  only  T^  of  $125  =  $1.25,  and  for  $4 
I  shall  have  4  times  $1.25  =  $  5,  answer  as  before. 

4.  If  I  buy  cloth  at  $  5  per  yard,  for  what  must  I  dispose  of 
it  per  yard  to  lose  20  per  cent.  ?  Ans.  $  4. 

OPERATION    BY   PROPORTION. 

$100  —  $  20  =  $  80  ;  $100  :  $  80  : :  $  5  :  $  4  Ans. 

By  analysis.  If  $  80  are  to  be  received  for  $100,  it  is  cer- 
tain "that  for  $  1  there  will  be  paid  only  T8a°Tr  =  fc  of  a  dollar  = 


SECT.  LVIII.]  PROFIT  AND  LOSS.  231 

$  0.80,  and  for  $  5  I  can  receive  only  5  times  $  0.80  =  $  4,  an- 
swer as  before  ;  therefore,  to  ascertain  at  what  price  to  sell  an 
article  to  gain  or  lose  a  certain  per  cent,  we  adopt  the  following 

RULE.  —  As  $  100  is  to  $  100  with  the  profit  added  or  loss  subtract- 
ed,  so  is  the  price  given  to  the  price  required. 

5.  If  I  sell  cloth  at  $  5  per  yard,  and  thereby  make  25  per 
cent.,  what  was  the  prime  cost  of  the  goods  ? 

OPERATION    B7    PROPORTION. 

$100  +  $25  =  $125;  $125  :  $100  ::  $5  :  $4   Ans. 

By  analysis.  As  $  125  are  received  for  $  100,  it  is  evident 
that  for  $1  there  will  be  received  only  |££  =  |  of  a  dollar  = 
$  0.80  ;  and  for  $  5,  5  times  $  0.80  =  $  4.00  Ans. 

6.  If  I  dispose  of  cloth  at  $  4  per  yard,  and  by  so  doing  lose 
20  per  cent.,  required  the  prime  cost  of  the  goods.     Ans.  $  5. 

OPERATION    BY    PROPORTION. 

$100  —  $20  =  $80;  $80  :  100  ::  $4  :  $ 5  Ans. 

By  analysis.  As  $100  are  received  for  $80,  it  is  certain 
that  for  J$p-  =  J  of  a  dollar  ==  $1.25  there  would  be  received 
only  $1  ;  therefore,  for  $4  which  I  receive  I  should  have  had 
4  times  $1.25  =  $5,  answer  as  before  ;  therefore,  when  we 
wish  to  ascertain  the  price  of  an  article,  when  we  know  what 
per  cent,  is  gained  or  lost,  we  adopt  the  following 

RULE.  —  As  $  WQ.ioith  the  gain  per  cent,  added  or  loss  per  cent,  sub- 
tracted is  to  $  100,  so  is  the  price  to  the  prime  cost. 

7.  If  I  sell  cloth  at  $  5  per  yard,  and  thereby  gain  25  per 
cent.,  what  would  be  my  gain  if  I  were  to  obtain  $  7  per  yard  ? 

Ans.  75  per  cent. 

OPERATION    BY    PROPORTION. 

$100 +  $25  =$125;  $5:  $7::  $125:  $175; 
$175  __  $100  =  $  75,  that  is,  75  per  cent. 

By  analysis.  As  $  5  amounts  to  $125,  it  is  evident  that  $  7 
will  amount  to  %  of  $125  =  $175  ;  and  if  $  7  amount  to  $175 
it  is  certain  that  $175  —  $100  =  $75  are  gained  on  $100, 
that  is,  75  per  cent. 

8.  If  I  purchase  cloth  at  $  7  per  yard,  and  thereby  gain  75 
per  cent.,  do  I  gain  or  lose  if  I  sell  the  same  at  $  3  per  yard  ? 

Ans.  lose  25  per  cent. 


232  PROFIT   AND   LOSS.  [SECT.  LVIII. 

OPERATION    BY    PROPORTION. 

$100 +$75  =  $175;  $7:  $3::  $175:  $75. 
$100  —  $  75  =  $25  ;  the  loss  is  therefore  25  per  cent. 

By  analysis.  If  $  7  give  $  175,  $  3  will  give  f  of  $175  = 
$75.  Therefore,  for  $100  there  are  received  only  $75 ;  there- 
fore there  is  $100  —  $  75  =  $  25,  or  25  per  cent,  loss,  answer. 

If,  therefore,  goods  be  sold  at  a  certain  price,  and  there  be 
gained  or  lost  a  certain  per  cent.,  and  we  wish  to  ascertain  what 
would  be  gained  or  lost  per  cent,  at  some  other  price,  we  de- 
duce the  following 

RULE.  — As  the  first  price  is  to  the  proposed  price,  so  is  $  100  with 
the  profit  per,  cent,  added  or  the  loss  per  cent,  subtracted  to  the  gain  or 
loss  per  cent,  at  the  assumed  price. 

NOTE. —  If  the  answer  exceeds  $100,  the  excess  is  the  gain  per  cent. ; 
but  if  It  be  less  than  $100,  the  deficiency  is  the  loss  per  cent. 

EXAMPLES    TO    EXERCISE   THE    PRECEDING   RULES. 

9.  Sold  broadcloth  at  $  6.12£  per  yard,  and  by  so  doing  lost 
12£  per  cent.     What  was  the  original  cost  per  yard  ? 

Ans.  $  7.00. 

By  analysis.  If  12£  per  cent,  be  lost,  87£  per  cent,  will  remain. 
It  is  now  required  to  find  of  what  number  $  6. 12£  is  j^.  This  is 

done  by  multiplying  $6.124-  by  100,  and  dividing  by  87£,  and 
it  produces  the  answer,  $  7X)0. 

10.  Bought  cloth  at  $  7.00  per  yard,  and  sold  it  at  $  6.12£. 
What  per  cent,  did  I  lose  ?  Ans.  12£  per  cent. 

11.  Bought  cloth  at  $7.00  per  yard,  and  sold  it  for  12£  per 
cent,  less  than  what  it  cost.     What  did  I  receive  ? 

Ans.  $6.12£. 

12.  Bought  cloth  at  $  3.60  per  yard.     For  how  much  must 
I  sell  it  to  gain  12£  per  cent.  ?  Ans.  $  4.05. 

13.  Sold  cloth  at  $  4.05  per  yard,  and  by  so  doing  I  gained 
124-  per  cent.     What  did  it  cost  ?  Ans.  $  3.60. 

14.  Bargained  for  cheese  at  $  8.50  per  cwt.     For  how  much 
must  I  sell  it  to  gain  10  per  cent.  ?         Ans.  $  9.35  per  cwt. 

15.  Sold  cheese  at  $  9.35  per  cwt.  and  gained  10  per  cent. 
What  did  I  give  for  it  ?  Ans.  $8.50  per  cwt. 

16.  Sold  cloth  at  $1.25  per  yard,  and  by  so  doing  lost  15  per 
cent.     For  what  should  I  have  sold  it  to  gain  12  per  cent.  ? 

Ans.  $1.64,7^  per  yard. 

17.  Sold  cloth  at  $1.25  per  yard,  and  lost  15  per  cent.   What 


SECT.  LVIII.]  PROFIT    AND  LOSS.  233 

per  cent,  should  I  have  gained  had  I  sold  it  for  $1.64,7^  per 
yard  ?  Ans.  12  per  cent. 

18.  Sold  cloth  for  $1.64,7Ty  per  yard,  and  gained  12  per 
cent.     For  what  should  I  have  sold  it  to  lose  15  per  cent.  ? 

Ans.  $1.25  per  yard. 

19.  Sold  cloth  for  $1.64,7T1T  per  yard,  and  gained    12  per 
cent.     What  should  I  have  lost  had  I  sold  it  for  $1.25  per 
yard  ?  Ans.  15  per  cent. 

20.  A  buys  corn  for  $  0.90  per  bushel,  and  sells  it  for  $1.20. 
B  buys  for  $1.12J,  and  sells  for  $1.50.     Who  gains  the  most 
per  cent.  ?  Ans.  both  gain  alike. 

21.  If  I  buy  cotton  cloth  at  13  cents  per  yard,  on  8  months' 
credit,  and  sell  it  again  at  12  cents  cash,  do  I  gain  or  lose,  and 
how  much  per  cent.  ?  Ans.  lose  4  per  cent. 

22.  If  24  yards  of  cloth  are  sold  at  $  2.50  per  yard,  and 
there  is  7£  per  cent,  loss  in  the  sale,  what  is  the  prime  cost  of 
the  whole  ?  Ans.  $  64.86,4ff . 

23.  Bought  24  yards  of  cloth  for  $  64.86,4f  f .     For  what 
must  I  sell  it  per  yard  to  lost  7£  per  cent.  ?          Ans.  $  2.50. 

24.  Bought  a  certain  quantity  of  cloth  for  $  64.86,4ff ,  and 
by  selling  it  at  $  2.50  per  yard,  I  lost  7|-  per  cent.     How  many 
yards  were  bought  ?  Ans.  24  yards. 

25.  Bought  24  yards  of  cloth  for  $  64.86,4f  f ,  and  sold  it  at 
$  2.50  per  yard  ;  what  per  cent,  is  lost  ?      Ans.  7£  per  cent. 

26.  If  27|cwt.  of  sugar  be  sold  at  $12.50  per  cwt.,  and  there 
is  gained  17  per  cent.,  what  was  the  first  cost  ? 

Ans.  $10.68,3T8T9T. 

27.  Sold  a  horse  for  $  75,  and  by  so  doing  I  lost  25  per 
cent.  ;  whereas,  I  ought  to  have   gained  30  per  cent.     How 
much  was  he  sold  under  his  real  value  ?  Ans.  $  55.00. 

28.  Bought  a  horse  which  was  worth  30  per  cent,  more  than 
I  gave  for  him  ;  but  having  been  injured,  I  sold  him  for  25  per 
cent,  less  than  what  he  cost,  and  thereby  lost  $  55  on  his  origi- 
nal value.  What  was  received  for  the  horse  ?     Ans.  $  75.00. 

29.  Bought  molasses  at  42  cents  per  gallon,  but  not  proving 
so   good  as  I  expected,  I  am  willing  to  lose  5  per  cent.     For 
what  must  I  sell  it  per  gallon  ?  Ans.  $  0.39,9. 

30.  Bought  a  hogshead  of  molasses  for  $112,  but  15  gallons 
having  leaked  out,  I  am  willing  to  lose  5  per  cent,  on  the  cost. 
For  how  much  per  gallon  must  I  sell  it  ?         Ans.  $2.21,6§. 

31.  Bought  a  hogshead  of  molasses  for  $112,  but  a  number 
of  gallons  having  leaked  out,  I  sell  the  remainder  for  $2.21,6| 

20* 


• 


234  DUODECIMALS.  [SECT.  LIX. 

per  gallon,  and  by  so  doing  I  lose  5  per  cent,  on  the  cost.    How 
many  gallons  leaked  out  ?  Ans.  15  gallons. 

32.  Bought  a  hogshead  of  molasses  for  a  certain  sum  ;  but 
15  gallons  having  leaked  out,  I  sell  the  remainder  for  $2.21,6$ 
per  gallon,  and  thereby  lose  5  per  cent,  on  the  cost.     What 
was  the  cost  ?  Ans.  $112.00. 

33.  Bought  a  hogshead  of  molasses  for  $112.00;  but  15  gal- 
Ions  having  leaked  out,  1  sell  the  remainder  at  $2.21,6§  per 
gallon.     What  per  cent,  is  my  loss  ?  Ans.  5  per  cent. 

34.  If  I  sell  cloth  at  $  5.60  per  yard,  and  thereby  lose  7  per 
cent.,  should  I  gain  or  lose,  and  how  much  per  cent,  by  selling 
it  at  $  6.25  per  yard  ?  Ans.  3r8/2-  per  cent.  gain. 

35.  Sold  a  watch  which  cost  me  $  30  for  $  35,  on  a  credit  of 
8  months.    What  did  I  gain  by  the  bargain  ?    Ans.  $  3.65,3|£. 

36.  When  tea  is  sold  at  $1.25  per  Ib.  there  is  lost  25  per 
cent. ;  what  would  be  the  gain  or  loss  per  cent,  if  it  should  be 
sold  at  $1.40  per  Ib.  ?  Ans.  16  per  cent.  loss. 

37.  A  exchanges  with  B  501bs.  of  indigo  at  $1.00  per  Ib. 
cash,  and  in  barter  $1.35  ;  but  he  is  willing  to  lose  12  per  cent, 
to  have  one  third  ready  money.     What  is  the  cash  price  of  1 
yard  of  cloth  delivered  by  B  at  $  5.00  per  yard  to  equal  A's 
bartering  price  reduced  12  per  cent.,  and  how  many  yards  were 
delivered  ?  Ans.  $  4.20§f f  cash  price  of  1  yard  ; 

7f  f  yards  delivered  by  B. 


SECTION  LIX. 
DUODECIMALS. 

DUODECIMALS  are  so  called  because  they  decrease  by  twelves 
from  the  place  of  feet  towards  the  right. 

Inches  are  called  primes,  and  are  marked  thus ' ;  the  next  di- 
vision after  is  called  seconds,  marked  thus  " ;  the  next  is  thirds, 
marked  thus  "'  ;  and  so  on. 

Duodecimals  are  commonly  used  by  workmen  and  artificers 
in  finding  the  contents  of  their  work. 

EXAMPLES. 

1.  Multiply  6  feet  8  inches  by  4  feet  5  inches. 


SECT.  LIX.]  DUODECIMALS.  235 

OPERATION.       As  feet  are  the  integers  or  units,  it  is  evident  that 
6  8'  feet  multiplied  by  feet  will  produce  feet ;  and  as 

4  5  inches  are  twelfths  of  a  foot,  the  product  of  inches 

2Q  g/  by  feet  will   be  twelfths  of  a  foot.     For  the  same 

29  4"  reason,  inches  multiplied  by  inches  will  produce 
twelfths  of  an  inch,  or  one  hundred  and  forty- 
fourths  of  a  foot. 

RULE.  —  Under  the  multiplicand  write  the  same  names  or  denomina- 
tions of  the  multiplier ;  that  is,  feet  under  feet,  inches  under  inches,  <5fc. 
Multiply  each  term  in  the  multiplicand,  beginning  at  the  lowest,  by  th? 
feet  of  the  multiplier,  and  write  each  result  under  its  respective  term,  ob- 
serving to  carry  a  unit  for  every  12  from  each  denomination  to  its  next 
superior.  In  the  same  manner  multiply  the  multiplicand  by  the  rnc/ies 
of  the  multiplier,  and  write  the,  result  of  each  term  one  place  further  to- 
wards the  right  than  the  corresponding  terms  in  tlie  preceding  product. 
Proceed  in  t/te  same  manner  with  the  seconds  and  all  the  rest  of  the  de- 
nominations, and  the  sum  of  the  several  products  will  be  the  product  re- 
quired. 

The  denomination  of  the  particular  products  will  be  as  follows. 

Feet  multiplied  by  feet  give  feet. 
Feet  multiplied  by  primes  give  primes. 
Feet  multiplied  by  seconds  give  seconds. 
Primes  multiplied  by  primes  give  seconds. 
Primes  .multiplied  by  seconds  give  thirds. 
Primes  multiplied  by  thirds  give  fourths. 
Seconds  multiplied  by  seconds  give  fourths. 
Seconds  multiplied  by  thirds  give  fifths. 
Seconds  multiplied  by  fourths  give  sixths. 
Thirds  multiplied  by  thirds  give  sixths. 
Thirds  multiplied  by  fourths  give  sevenths. 
Thirds  multiplied  by  fifths  give  eighths,  &c. 

2.  Multiply  4ft.  7'  by  6ft.  4'.  Ans.  29ft.  0'  4". 

3.  Multiply  14ft.  9'  by  4ft.  6'.  Ans.  66ft.  4'  6". 

4.  Multiply  4ft.  7'  8"  by  9ft.  6'.  Ans.  44ft.  0'  10". 

5.  Multiply  10ft.  4'  5"  by  7ft.  8'  6". 

Ans.  79ft.  11'  0"  6"'  6"". 

6.  Multiply  39ft.  10'  7"  by  18ft.  8'  4". 

Ans.  745ft.  6'  10"  2'"  4"". 

7.  How  many  square  feet  in  a  floor  48  feet  6  inches  long,  24 
feet  3  inches  broad  ?  Ans.  1176ft.  1'  6". 

8.  What  are  the  contents  of  a  marble  slab,  whose  length  is  5 
feet  7  inches,  and  breadth  1  foot  10  inches  ? 

Ans.  10ft.  2'  10". 


236  DUODECIMALS.  [SECT.  ux. 

9.  The  length  of  a  room  being  20  feet,  its  breadth  14  feet 
6  inches,  and  height   10  feet  4  inches,  how  many  yards  of 
painting  are  in  it,  deducting  a  surplus  of  4  feet  by  4  feet  4 
inches,  and  2  windows,  each  6  feet  by  3  feet  2  inches  ? 

Ans.  7322T  yards. 

10.  Required  the  solid  contents  of  a  wall  53  feet  6  inches 
long,  10  feet  3  inches  high,  and  2  feet  thick. 

Ans.  1096ft.  9'. 

11.  There  is  a  house  with  four  tiers  of  windows,  and  4  win- 
dows in  a  tier  ;  the  height  of  the   first  is  6  feet  8  inches  ;  the 
second,  5  feet  9  inches  ;  the  third,  4  feet  6  inches  ;  the  fourth,  3 
feet  10  inches  ;  and  the  breadth  is  3  feet  5  inches  ;  how  many 
square  feet  do  they  contain  in  the  whole  ?       Ans.  283ft.  7in. 

12.  How  many  feet  of  boards  would  it  require  to  make  15 
boxes,  each  of  which  is  7  feet  9  inches  long,  3  feet  4  inches 
wide,  and  2  feet   10  inches  high  ;  and  how  many  cubic  yards 
would  they  contain  ?          Ans.  1717ft.  lin.  40f^|  cubic  yds. 

13.  A  mason  has  plastered  3  rooms ;  the  ceiling  of  each  is 
20  feet  by  16  feet  6  inches,  the  walls  of  each  are  9  feet  6  inches 
high,  and  there  are  to  be  90  yards  deducted  for  doors,  windows, 
&c.     How  many  yards  must  he  be  paid  for  ? 

Ans.  251yd.  1ft.  6in. 

14.  How  many  feet  in  a  board  which  is  17  feet  6  inches  long, 
and  1  foot  7  inches  wide  ?  Ans.  27ft.  8'  6". 

15.  How  many  feet  in  a  board  27  feet  9  inches  long,  29 
inches  wide  ?  Ans.  67ft.  0'  9". 

16.  How  many  feet  of  boards  will  it  take  to  cover  the  side 
of  a  building  47  feet  long,  17  feet  9  inches  high  ? 

Ans.  834ft.  3'. 

NDTE.  —  A  board  to  be  merchantable  should  be  1  inch  thick ;  therefore 
to  reduce  a  plank  to  board  measure,  the  superficial  contents  of  the  plank 
should  be  multiplied  by  its  thickness. 

17.  How  many  feet,  board  measure,  are  in  a  plank  18  feet 
9  inches  long,  1  foot  6  inches  wide,  and  3  inches  thick  ? 

Ans.  84ft.  4'  6". 

18.  How  many  feet,  board  measure,  are  in  a  plank  20  feet 
long,  1  foot  6  inches  wide,  and  2£  inches  thick  ?     Ans.  75ft. 

19.  How  many  feet  in  a  plank  40  feet  6  inches  long,  30 
inches  wide,  and  2f  inches  thick  ?  Ans.  278ft.  5'  3". 

NOTE.  —  A  pile  of  wood  that  is  8  feet  long,  4  feet  high,  and  4  feet  wide, 
contains  128  cubic  feet,  or  a  cord,  and  every  cord  contains  8  cord-feet;  and 
as  8  is  ^  of  128,  every  cord-foot  contains  16  cubic  feet ;  therefore,  divid- 
ing the  cubic  feet  in  a  pile  of  wood  by  16,  the  quotient  is  the  cord  feet ; 
and  if  cord-feet  be  divided  by  8,  the  quotient  is  cords. 


SECT.  Lix.J  DUODECIMALS.  237 

20.  How  many  cords  of  wood  in  a  pile  18  feet  long,  6  feet 
high,  and  4  feet  wide  ?  Ans.  3|  cords. 

21.  How  many  cords  in  a  pile   10  feet  long,  5  feet  high,  7 
feet  wide  ?  Ans.  2  cords,  94  cubic  feet. 

22.  How  many  cords  in  a  pile  35  feet  long,  4  feet  wide,  4 
feet,  high  ?  Ans.  4f  cords. 

23.  How  many  cords  in  a  pile  that  measures  8  feet  on  each 
side  ?  Ans.  4  cords. 

24.  How  many  cords  in  a  pile  that  is  10  feet  on  each  side  ? 

Ans.  7|f  cords. 

.NOTE.  —  When  wood  is  "  corded  "  in  a  pile  4  feet  wide,  by  multiply- 
ing its  length  by  its  height,  and  dividing  the  product  by  4,  the  quotient  is 
the  cord-feet ;  and  if  a  load  of  wood  be  8  feet  long,  and  its  height  be  mul- 
tiplied by  its  width,  and  the  product  divided  by  2,  the  quotient  is  the 
cord-feet. 

25.  How  many  cords  of  wood  in  a  pile  4  feet  wide,  70  feet 
6  inches  long,  and  5  feet  3  inches  high  ?      Ans.  11T9^  cords. 

NOTE.  —  Small  fractions  are  rejected. 

26.  How  many  cords  in  a  pile  of  wood  97  feet  9  inches  long* 
4  feet  wide,  and  3  feet  6  inches  high  ?        Ans.  10£££  cords. 

27.  Required  the  number  of  cords  of  wood  in  a  pile  100  feet 
long,  4  feet  wide,  and  6  feet  11  inches  high.          Ans.  21ff. 

28.  Agreed  with  a  man  for  10  cords  of  wood,  at  $  5.00  a 
cord  ;  it  was  to  be  cut  4  feet  long,  but  by  mistake  it  was  cut  on- 
ly 46  inches  long.     How  much  injustice  should  be  deducted 
from  the  stipulated  price  ?  Ans.  $  2.08£. 

29.  If  a  load  of  wood  be  8  feet  long,  3  feet  8  inches  wide, 
and  5  feet  high,  how  much  does  it  contain  ? 

Ans.  9£  cord-feet. 

30.  If  a  load  of  wood  be  8  feet  long,  3  feet  10  inches  wide, 
and  6  feet  6  inches  high,  how  much  does  it  contain  ? 

Ans.  124-£  cord-feet. 

31.  If  a  load  of  wood  be  8  feet  long,  3  feet  6  inches  wide, 
how  high  should  it  be  to  contain  1  cord  ?      Ans.  4ft.  6   lOf '. 

32.  If  a  load  of  wood  be  12  feet  long,  and  3  feet  9  inches 
wide,  how  high  should  it  be  to  contain  2  cords  ? 

Ans.  5ft.  8'  3£". 

33.  D.   H.   Sanborn's   parlour  is  17ft.  9in.  long,  14ft.  Sin. 
wide,  and  8ft.  9in.  high.     There  are  two  doors  3ft.  4in.  wide, 
and   7ft.  high,  and  four  windows  5ft.  3in.  high,  and  3ft.  4in. 
wide  ;  the  mop-boards  are  9in.  high.     B.  Gordon,  a  first-rate 
mason,  will  charge  10  cents  per  square  yard  for  plastering  the 


238  INVOLUTION.  [SECT.  LX. 

room.  The  paper  for  the  room  is  20  inches  wide,  and  costs 
6£  cents  per  yard.  E.  Eaton  will  "  paper  "  the  room  for  4 
cents  per  square  yard.  Each  window  has  12  lights  of  lOin.  by 
14in.  glass,  the  price  of  which  is  12£  cents  per  square  foot. 
The  painter's  bill  for  setting  the  glass  is  8  cents  per  light,  and 
for  painting  the  floor,  mop-boards,  and  doors  is  25  cents  per 
square  yard.  What  is  the  amount  of  Mr.  Sanborn's  bill  ? 

Ans.  $  33.72/TV 


SECTION  LX. 
INVOLUTION. 

INVOLUTION  is  the  raising  of  powers  from  any  given  number, 
as  a  root. 

A  power  is  a  quantity  produced  by  multiplying  any  given 
number,  called  a  root,  a  certain  number  of  times  continually  by 
itself ;  thus, 

2  =   2  is  the  root,  or  1st  power  of  2  =  21. 

2x2=    4  is  the  2d  power,  or  square  of        2  =  22. 

2x2x2=    8  is  the  3d  power,  or  cube  of  2  =  23. 

2  X  2  X  2  X  2  =  16  is  the  4th  power,  or  biquadrate  of  2  =  24. 

The  number  denoting  the  power  is  called  the  index  or  expo- 
nent of  the  power.  Thus,  the  fourth  power  of  3,  =  81,  is  ex- 
pressed by  34,  and  4  is  the  index  or  exponent ;  and  the  second 
power  of  7,  —  49,  is  expressed  by  72. 

To  raise  a  number  to  any  power  required. 

RULE.  —  Multiply  the  given  number  continually  by  itself,  till  the 
number  of  multiplications  be  one  less  than  the  index  of  the  power  to  be 
found,  and  the  last  product  will  be  the  power  required. 

EXAMPLES. 

1.  What  is  the  5th  power  of  4  ? 

4x4x4x4x4=  1024  Ans. 

2.  What  is  the  3d  power  of  8  ?  Ans.  512. 

3.  What  is  the  10th  power  of  7  ?  Ans.  282475249. 

4.  What  is  the  6th  power  of  5  ?  Ans.  15625. 

5.  What  is  the  3d  power  of  £  ?  Ans.  ££. 

6.  What  is  the  5th  power  of  £  ?  Ans.  r^. 

7.  What  is  the  4th  power  of  2f  ?  Ans.  50|f . 


SECT.  LX.]  EVOLUTION.  239 

8.  What  is  the  6th  power  of  If  ?  Ans.  16|f  tti« 

9.  What  is  the  4th  power  of  .045  ?    Ans.  .000004100625. 
10.  What  is  the  0  power  of  1728  ?  Ans.  1. 


EVOLUTION, 
OR  THE  EXTRACTION  OF  ROOTS. 

EVOLUTION  is  the  reverse  of  Involution,  and  teaches  to  find 
the  roots  of  any  given  powers. 

The  root  is  a  number  whose  continual  multiplication  into  it- 
self produces  the  power,  which  is  denominated  the  2d,  3d,  4th, 
&c.,  power,  according  to  the  number  of  times  which  the  root  is 
multiplied  into  itself.  Thus,  4  is  the  square  root  of  16,  because 
4  x  4  z=  16  ;  and  3  is  the  cube  root  of  27,  because  3x3x3 
—  27  ;  and  so  on. 

Although  there  is  no  number  of  which  we  cannot  find  any 
power  exactly,  yet  there  are  many  numbers  of  which  precise 
roots  can  never  be  determined  ;  but,  by  the  help  of  decimals, 
we  can  approximate  towards  the  root  to  any  assigned  degree  of 
exactness. 

The  roots  which  approximate  are  called  surd  roots ;  and  those 
which  are  perfectly  accurate  are  called  rational  roots. 

Roots  are  sometimes  denoted  by  writing  the  character  x/  be- 
fore the  power,  with  the  index  of  the  root  over  it ;  thus,  the  3d 
root  of  36  is  expressed  \/36,  and  the  second  root  of  36  is  <,/ 
36,  the  index  2  being  omitted  when  the  square  root  is  designed. 

If  the  power  be  expressed  by  several  numbers  with  the  sign 
-f-  or  —  between  them,  a  line  is  drawn  from  the  top  of  the  sign 

over  all  the  parts  of  it ;  thus,  the  3d  root  of  42+22  is  ^/42+22, 

and  the  second  root  of  59  —  17  is  v/59  —  17,  &c. 

Sometimes  roots  are  designated  like  powers  with   fractional 

indices.  Thus  the  square  root  of  15  is  152",  the  cube  root  of  21 
is  21*,  and  the  4th  root  of  37  —  20  is  37  —  20^,  &c. 

It  sometimes  will  happen  that  one  root  is  involved  in  another, 
thus: 


V  125  —  5  -f  x/19  4-  6,  or  v/161  — 


#178  +  7  y^Ts  +  ^84^5  _  ^/87  +  16. 


240 


TABLE   OF  POWERS. 


[SKCT.  LX. 


05 


00 


c* 


00 


3 


CO 


00 


O 


9 


CO 


CO 


679616 


tft 


CO 


CO 


CO 


s 


CO 


CO 


(^ 


co 


05 


co 


CO 


05 


<?* 


00 


CO 


.3 


s 


I 
3 


i* 

o 

I 


0) 

I 


SECT.  LXI.]  SQUARE   ROOT.  241 

SECTION  LXI. 

EXTRACTION  OF  THE  SQUARE  ROOT. 

1.  LET  it  be  required  to  find  what  number  multiplied  into  it- 
self will  produce  1296. 

OPERATION.  In  contemplating  this 

1296(30  +  6  =  36  Ans.     problem,   we    perceive 

900  that  the  root  or  number 

M    ,    ,*       g/^xoru?  sought  must  consist  of 

60  +  6 •  =  66)396  ^  figureg)  since  the 

product    of    any    two 

numbers  can  have  at  most  but  as  many  figures  as  there  are  in 
both  factors,  and  at  least,  but  one  less.  We  perceive,  also,  that 
the  first  figure  of  the  root  multiplied  by  itself  must  give  a  num- 
ber not  exceeding  12,  and  as  12  is  not  the  second  power  of  any 
number,  and  the  second  power  of  4  is  more  than  12,  we  take  3 
for  the  first  figure  of  the  root,  which  multiplied  into  itself  gives 
9.  Now,  the  second  power  of  3,  considered  as  occupying  the 
place  of  tens,  is  900,  of  which  we  have  the  root  30.  Taking 
900  from  1296,  we  have  a  remainder,  396 ;  and  having  found 
the  root  of  900,  we  are  now  to  seek  a  number,  which,  being 
added  to  this  root  (30)  and  multiplied  into  itself  once,  and  into 
30  twice*  will  produce  396.  This  number  is  found  by  dividing 
396  by  twice  30  plus  the  number  sought.  Q.  E.  D. 

NOTE.  —  Owing  to  the  fact  that  the  number  of  figures  in  the  product  of 
any  two  numbers  is  always  limited  as  above  stated,  we  ascertain  the  num- 
ber of  figures  in  the  root  of  any  given  second  power  by  putting  a  dot 
over  the  place  of  units,  then  over  the  place  of  hundreds,  and  so  on.  The 
number  of  dots  gives  the  number  of  figures  in  the  root.  Thus  the  square 
root  of  133225  consists  of  three  figures. 

2.  What  is  the  square  root  of  576  ? 

OPERATION.  To  illustrate   this  question   in  a  different 

576(24  Ans      way  from  the  first,  we  will  suppose  that  we 

400  have  576  tiles,  each  of  which  is  one  foot 

-— —  square,  and  we  wish  to  know  the  side  of  a 

14)176  square  room  whose  floor  they  will  pave  or 

cover. 

*  By  adding  6  to  30  and  multiplying  the  sum  (36)  into  itself,  we  can 
easily  see  that  we  multiply  6  by  itself  once,  and  30  by  6  twice,  since 
30  is  contained  in  both  factors,  and  in  the  operation  is  multiplied  by  the 
6  in  each. 

21 


242  SQUARE   ROOT.  [SECT.  LXI. 

If  we  find  a  number  which  multiplied  into  itself  will  produce 
576,  that  number  will  give  the  side  of  the  room  required.  We 
perceive  that  as  our  number  (576)  consists  of  three  figures, 
there  will  be  two  figures  in  its  root,  since  the  square  of  no  num- 
ber expressed  by  a  single  figure  can  be  so  large  as  576  ;  and, 
if  the  root  were  supposed  to  have  more  than  two  figures,  its 
square  would  exceed  576.  Dividing  the  number  into  periods 
thus,  576,  we  now  find  by  trial,  or  by  the  table  of  powers,  that 
the  greatest  square  number  or  second  power  in  the  left-hand 
period,  5  (hundred),  is  4  (hundred),  and  that  its  root  is  2, 
which  we  write  in  the  quotient.  (See  operation.)  As  this  2  is 
in  the  place  of  tens,  its  value  must  be  20,  and  its  square  or  sec- 
ond power  400. 

Let   this   be  represented   by  a  square 
whose  sides  measure  20  feet  each,  and  Fig.  1. 

whose    contents   will    therefore    be   400  20 

square  feet.    See  figure  D.    We  now  sub- 


D 

20 
20 
400 


tract  400  from  576  and  there  remain  176 

square  feet  to  be  arranged  on  two  sides  of 

the  figure  D,  in  order  that  its  form  may 

remain  square.     We  therefore  double  the 

root  20,  one  of  the  sides,  and  it  gives  the 

length  of  the  two  sides  to  be  enlarged, 

viz.  40.       We   then   inquire    how  many 

times  40  as  a  divisor  is  contained  in  the  dividend  (except  the 

right-hand  figure)  and  find  it  to  be  4  times :  this  we  write  in 

the  root,  and  also  in  the  divisor. 

This  4  is  the  breadth  of  the  addi- 
tion to  our  square.     (See  figure  2.)  Fig.  2. 

And  this  breadth,  multiplied  by  the       20_ 

length  of  the  two  additions  (40),  gives 
the  contents  of  the  two  figures  E  and 
F,  160  square  feet,  which  is  80  feet 


D 

20 
20 

400 


F 

20 
4 

80 


for  each. 

There  now  remains  the  space  G,  to 
complete  the  square,  each  side  of 
which  is  4  feet ;  it  being  equal  to  the 
breadth  of  the  additions  E  and  F. 
Therefore,  if  we  square  4  we  have 
the  contents  of  the  last  addition  G  = 

16.  It  is  on  account  of  this  last  addition  that  the  last  figure  of 
the  root  is  placed  in  the  divisor.  If  we  now  multiply  the  divi- 
sor, 44,  by  the  last  figure  in  the  root  (4),  the  product  will  be 


SECT.  LSI.]  SQUARE  ROOT.  243 

176,  which  is  equal  to  the  remain-  D  contains  400  square  feet. 

ing  feet  after  we  had  formed  our  E       "          80      "         " 

first  square,  and  equal  to  the  addi-  F       "          80      "         " 

tions  E,  F,  and  G,  in  figure  2.  G       "          16      "        " 

We  therefore  perceive  that  fig-  Proof  576 

ure  2  may  represent  a  floor  24  or 

feet  square,  containing  576  square  34  ^  24= 576 

feet. 

RULE.  —  1 .  Distinguish  the  given  number  into  periods  of  two  figures 
each,  by  putting  a  point  over  the  place  of  units,  another  over  the  place  of 
hundreds,  and  so  on,  which  points  show  the  number  of  figures  the  root 
will  consist  of. 

2.  Find  the  greatest  square  number  in  the  first  or  left-hand  period, 
placing  the  root  of  it  at  the  right  hand  of  the  given  number  (after  the 
manner  of  a  quotient  in  division),  for  the  first  figure  of  the  root,  and 
the  square  number  under  the  period,  subtracting  it  therefrom;  and  to 
the  remainder  bring  down  the  next  "period  for  a  dividend,  always,  how- 
ever, omitting  the  right-hand  figure  of  this  dividend  in  dividing. 

3.  Place  the  double  of  the  root  already  found  on  the  left  hand  of  the 
dividend  for  a  divisor. 

4.  Find  how  often  the  divisor  is  contained  in  the  dividend  (omitting 
the  right-hand  figure) ,  placing  the  answer  in  the  root  for  tfte  second  fig- 
ure of  it,  and  likewise  on  the  right  hand  of  the  divisor.*     Multiply  the 
divisor  with  the  figure  last  annexed  by  the  figure  last  placed  in  the  root, 
and  subtract  the  product  from  the  dividend.     To  the  remainder  join  the 
next  period  for  a  new  dividend. 

5.  Double  the  figures  already  found  in  the  root  for  a  new  divisor  (or 
bring  down  the  last  divisor  for  a  new  one,  doubling  the  right-hand  fig- 
ure of  it),  and  from  these  find  the  next  figure  in  the  root  as  last  directed, 
and  continue  the  operation  in  the  same  manner,  till  you  have  brought 
down  all  the  periods. 


NOTE  1.  —  If,  when  the  given  power  is  pointed  off  as  the  case  requires, 
the  left-hand  period  should  be  deficient,  it  must  nevertheless  stand  as  the 
first  period. 

NOTE  2.  —  If  there  be  decimals  in  the  given  number,  it  must  be  pointed 
both  ways  from  the  place  of  units.  If,  when  there  are  integers,  the  first 
period  in  the  decimals  be  deficient,  it  may  be  completed  by  annexing  so 
many  ciphers  as  the  power  requires.  And  the  root  must  be  made  to 
consist  of  so  many  whole  numbers  and  decimals  as  there  are  periods  be- 
longing to  each  ;  and  when  the  periods  belonging  to  the  given  numbers 
are  exhausted,  the  operation  may  be  continued  at  pleasure  by  annexing 
ciphers. 


*  One  or  two  units  are  generally  to  be  allowed  on  account  of  other  de- 
ficiencies in  enlarging  the  square. 


244  SQUARE  ROOT.  [SECT.  LXI. 

3.  What  is  the  square  root  of  278784  ? 

278784(528  Answer. 
25 

102)287 
204 


1048)8384 

8384 

4.  What  is  the  square  root  of  776161  ?  Ans.  881. 

5.  What  is  the  square  root  of  998001  ?  Ans.  999. 

6.  What  is  the  square  root  of  10342656  ?        Ans.  3216. 

7.  What  is  the  square  root  of  9645192360241  ? 

Ans.  3105671. 

8.  Extract  the  square  root  of  234.09.  Ans.  15.3. 

9.  Extract  the  square  root  of  .000729.  Ans.  .027. 

10.  Required  the  square  root  of  17.3056.  Ans.  4.16. 

11.  Required  the  square  root  of  373.  Ans.  19.3132079+. 

12.  Required  the  square  root  of  8.93.  Ans.  2.9883 1055-J-. 

13.  What  is  the  square  root  of  1.96  ?  Ans.  1.4. 

14.  Extract  the  square  root  of  3.15.    Ans.  1.77482393+. 

15.  What  is  the  square  root  of  572199960721  r 

Ans.  756439.. 

If  it  be  required  to  extract  the  square  root  of  a  vulgar  frac- 
tion, reduce  the  fraction  to  its  lowest  terms ;  then  extract  the 
square  root  of  the  numerator  for  a  new  numerator,  and  of  the 
denominator  for  a  new  denominator ;  or  reduce  the  vulgar  frac- 
tion to  a  decimal,  and  extract  its  root. 

16.  What  is  the  square  root  of  -rH^  ?  Ans.  -fa. 

17.  What  is  the  square  root  of  f  f  $f  ?  Ans.  T7g-. 

18.  What  is  the  square  root  of  42£  ?  Ans.  6£. 

19.  What  is  the  square  root  of  52T9^  ?  Ans.  7£. 

20.  What  is  the  square  root  of  95-^  ?  Ans.  9f . 

21.  What  is  the  square  root  of  363j£T  ?  Ans.  19TV 

22.  Extract  the  square  root  of  6f .  Ans.  2.5298-J-. 

23.  Extract  the  square  root  of  8|.  Ans.  2.9519-f-- 

24.  Required  the  square  root  of  2.  Ans.  1.41421-J-. 


APPLICATION  OF  THE   SQUARE   ROOT. 

DEFINITIONS. 

1.  A  circle  is  a  figure  bounded  by  a  line  equally  distant  from 
a  point  called  the  centre. 


SECT.  LXI.] 


SQUARE  ROOT. 


245 


2.  A  triangle  is  a  figure  of  three  sides. 

3.  An  equilateral  triangle  is  that  whose  three  sides  are  equal. 

4.  An  isosceles  triangle  is  that  which  has  two  sides  equal,  and 
a  perpendicular  from  the  angle  where  the  equal  sides  meet  bi- 
sects the  base. 

5.  A  right-angled  triangle  is  a  figure  of  three  sides  and  three 
angles,  one  of  which  is  a  right  angle.     See  figure  1. 

The  longest  side  is  called  the 
hypothenuse,  the  horizontal  side 
the  base,  and  the  other  side  is 
called  the  perpendicular. 


The  following  propositions  are  demonstrably  true. 


In  a  right-angled  tri- 
angle, the  square  of  the 
longest  side  is  equal  to 
the  sum  of  the  squares 
of  the  other  two  sides. 

In  all  similar  trian- 
gles, that  is,  in  all  trian- 
gles whose  correspond- 
ing angles  are  equal,  the 
sides  about  the  equal 
angles  are  in  direct  pro- 
portion to  each  other; 
that  is,  as  the  longest  side 
of  one  triangle  is  to  the 
longest  side  of  the  other, 
so  is  either  of  the  other 
sides  of  the  former  trian- 
gle to  the  corresponding 
side  of  the  latter. 

Let  ABC  and  DEF  be 
two  similar  triangles,  and 
let  A  B  be  6  feet,  B  C,  8 
feet,  and  A  C,  10  feet. 
Again,  let  D  E  be  12  feet, 
E  F,  16  feet,  and  D  F  will 
be  20  feet.  That  is,  B  C 
will  be  to  E  F  as  A  B  to 
D  E.  Then  8  feet  :  16 
21* 


Fig.  2. 


Fig.  3. 


Fig.  4. 


16 


12 


246  SQUARE  ROOT.  [SECT.  LXI. 

feet  : :  6  feet  :  12  feet.  Again,  A  B  will  be  to  D  E  as  A  C  to 
D  F.  That  is,  6  feet  :  12  feet  : :  10  feet  :  20  feet. 

Circles  are  to  each  other  as  the  squares  of  their  diameters. 

If  the  diameter  of  a  circle  be  multiplied  by  3.14159,  the 
product  is  the  circumference. 

If  the  square  of  the  diameter  of  a  circle  be  multiplied  by 
.785398,  the  product  is  the  area. 

If  the  square  root  of  half  the  square  of  the  diameter  of  a 
circle  be  extracted,  it  is  the  side  of  an  inscribed  square. 

If  the  area  of  a  circle  be  divided  by  .785398,  the  quotient  is 
the  square  of  the  diameter. 

EXAMPLES. 

25.  A  certain  general  has  an  army  of  141376  men.     How 
many  must  he  place  in  rank  and  file  to  form  them  into  a 
square  ?  Ans.  376. 

26.  If  the  area  of  a  circle  be   1760  yards,  how  many  feet 
must  the  side  of  a  square  measure  to  contain  that  quantity  ? 

Ans.  125.857+  feet. 

27.  If  the  diameter  of  a  round  stick  of  timber  be  24  inches, 
how  large  a  square  stick  may  be  hewn  from  it  ? 

Ans.  16.97+  inches. 

28.  I  wish  to  set  out  an  orchard  of  2400  mulberry-trees,  so 
that  the  length  shall  be  to  the  breadth  as  3  to  2,  and  the  distance 
of  each  tree,  one  from  the  other,  7  yards.     How  many  trees 
must  there  be  in  the  length,  and  how  many  in  the  breadth  ;  and 
on  how  many  square  yards  of  ground  will  they  stand  ? 

Ans.  60  in  length  ;  40  in  breadth  ;  112749  square  yards. 

29.  If  a  lead  pipe  £  of  an  inch  in  diameter  will  fill  a  cistern 
in  3  hours,  what  should  be  its  diameter  to  fill  it  in  2  hours  ? 

Ans.  .918+  inches. 

30.  If  a  pipe  1£  inches  in  diameter  will  fill  a  cistern  in  50 
minutes,  how  long  would  it  require  a  pipe  that  is  2  inches  in  di- 
ameter to  fill  the  same  cistern  ?  Ans.  28m.  7^-sec. 

31.  If  a  pipe  6  inches  in  diameter  will  draw  off  a  certain 
quantity  of  water  in  4  hours,  in  what  time  would  it  take  3  pipes 
of  four  inches  in  diameter  to  draw  off  twice  the  quantity  ? 

Ans.  6  hours. 

32.  If  a  line  144  feet  long  will  reach  from  the  top  of  a  fort 
to  the  opposite  side  of  a  river  that  is  64  feet  wide,  what  is  the 
height  of  the  fort  ?  Ans.  128.99+. 

33.  A  certain  room  is  20  feet  long,  16  feet  wide,  and  12  feet 
high  ;  how  long  must  a  line  be  to  extend  from  one  of  the  lower 
corners  to  an  opposite  upper  corner  ?  Ans.  28.28  feet. 


SECT.  LXI  ]  SQUARE  ROOT.  347 

34.  Two  ships  sail  from  the  same  port ;  one  goes  due  north 
128  miles,  the  other  due  east  72  miles ;  how  far  are  the  ships 
from  each  other  ?  Ans.  146.86+. 

35.  There  are  two  columns  in  the  ruins  of  Persepolis  left 
standing  upright ;  one  is  70  feet  above  the  plane,  and  the  other 
50  ;  in  a  straight  line  between  these  stands  a  small  statue,  5 
feet  in  height,  the  head  of  which  is  100  feet  from  the  summit 
of  the  higher,  and  80  feet  from  the  top  of  the  lower  column. 
Required  the  distance  between  the  tops  of  the  two  columns. 

Ans.  143.537+  feet. 

36.  The  height  of  a  tree,  growing  in  the  centre  of  a  circular 
island,  100  feet  in  diameter,  is   160  feet ;  and  a  line  extending 
from  the  top  of  it  to  the  farther  shore  is  400  feet.     What  is  the 
breadth  of  the  stream,  provided  the  land  on  each  side  of  the 
water  be  level  ?  Ans.  316.6+  feet. 

37.  A  ladder  70  feet  long  is  so  planted  as  to  reach  a  window 
40  feet  from  the  ground,  on  one  side  of  the  street,  and  without 
moving  it  at  the  foot  it  will  reach  a  window  30  feet  high  on  the 
other  side  ;  what  is  the  breadth  of  the  street  ? 

Ans.  120.69+  feet. 

38.  If  an  iron  wire  -^  inch  in  diameter  will  sustain  a  weight 
of  450  pounds,  what  weight  might  be  sustained  by  a  wire  an 
inch  in  diameter  ?  Ans.  45,000lbs. 

39.  A  tree  80  feet  in  height  stands  on  a  horizontal  plane  ;  at 
what  height  from  the  ground  must  it  be  cut  off,  so  that  the  top 
of  it  may  fall  on  a  point  40  feet  from  the  bottom  of  the  tree, 
the  end  where  it  was  cut  off  resting  on  the  stump  ? 

Ans.  30  feet. 

40.  Four  men,  A,  B,  C,  D,  bought  a  grindstone,  the  diameter 
of  which  was  4  feet ;  they  agreed  that  A  should  grind  off  his 
share  first,  and  that  each  man  should  have  it  alternately  until  he 
had  worn  off  his  share  ;  how  much  will  each  man  grind  off? 

Ans.  A  3.22+,  B  3.81+,  C  4.97+,  D  12  inches. 

41.  What  is  the  length  of  a  rope  that  must  be  tied  to  a  horse's 
neck,  that  he  may  feed  over  an  acre  ?       Ans.  7.136+  rods. 

42.  Required  the  greatest  possible  number  of  hills  of  corn 
that  can  be  planted  on  a  square  acre,  the  hills  to  occupy  only  a 
mathematical  point,  and  no  two  hills  to  be  within  three  and  a 
half  feet  of  each  other.  Ans.  4165. 

43.  James  Page  has  a  circular  garden,  10  rods  in  diameter  ; 
how  many  trees  ean  be  set  upon  it,  so  that  no  two  shall  be  with- 
in ten  feet  of  each  other,  and  no  tree  within  two  and  a  half  feet 
of  the  fence  inclosing  the  garden  ?  Ans.  241. 


248  CUBE   ROOT.  [SECT.  LXII. 

44.  A  gentleman  has  a  garden  in  the  form  of  an  equilateral 
triangle,  the  sides  whereof  are  200  feet.  At  each  corner  of 
the  garden  stands  a  tower ;  the  height  of  the  first  tower  is  30 
feet,  the  height  of  the  second,  40  feet,  and  the  height  of  the 
third,  50  feet.  At  what  distance  from  the  bottom  of  each  of 
these  towers  must  a  ladder  be  placed,  that  it  may  just  reach  the 
top  of  each  tower  ?  and  what  will  be  the  length  of  the  ladder, 
the  ground  of  the  garden  being  horizontal  ? 

Ans.  The  foot  of  the  ladder  from  the  base  of  the  first  tower, 
118.811+  feet;  second  tower,  115.827+  feet;  third  tower, 
111.875+  feet.  Length  of  the  ladder,  122.535+  feet. 


SECTION  LXII. 

EXTRACTION  OF  THE  CUBE  ROOT. 

A  CUBE  is  a  square  prism,  being  bounded  by  six  equal  side<, 
which  are  perpendicular  to  each  other. 

A  number  is  said  to  be  cubed  when  it  is  multiplied  into  its 
square. 

To  extract  the  cube  root  is  to  find  a  number,  which,  multi- 
plied into  its  square,  will  produce  the  given  number. 

RULE.  —  1.  Separate  the  given  number  into  periods  of  three  figures 
each,  by  putting  a  point  over  tJie  unit  figure,  and  every  third  figure  be- 
yond the  place  of  units. 

2.  Find  by  the  table  the  greatest  cube  in  the  left-hand  period,  and  put 
its  root  in  the  quotient. 

3.  Subtract  the  cube  thus  found  from  this  period,  and  to  the  remain- 
der bring  down  the  next  period  ;  call  this  the  dividend. 

4.  Multiply  the  square  of  the  quotient  by  300,  calling  it  tfie  triple 
square ;  multiply  also  the  quotient  by  30,  calling  it  the  triple  quotient ; 
the  sum  of  these  call  the  divisor. 

5.  Find  how  many  times  the  divisor  is  contained  in  the  dividend,  and 
place  the  result  in  the  quotient. 

NOTE.  —  One  or  two  units,  and  sometimes  three,  must  be  allowed. 

6.  Multiply  the  triple  square  by  the  last  quotient  figure,  and  write  the 
product  under  the  dividend ;  multiply  the  square  of  the  last  quotient  fig- 
ure by  the  triple  quotient,  and  place  this  product  under  the  last ;  under 
all,  set  the  cube  of  the  last  quotient  figure,  and  call  their  sum  the  subtra- 
hend. 


SECT.  LXIl.j 


CUBE  ROOT. 


249 


7.  Subtract  the  subtrahend  from  the  dividend,  and  to  the  remainder 
bring  down  the  next  period  for  a  new  dividend,  with  which  proceed  as 
before,  and  so  on,  till  the  whole  is  completed. 

NOTE.  —  The  same  rule  must  be  observed  for  continuing  the  operation, 
and  pointing  for  decimals,  as  in  the  square  root. 

ILLUSTRATION. 

We  suppose  we  have  46,656  cubic  blocks  of  granite,  which 
measure  one  foot  on  each  side.  WHh  these  we  wish  to  erect  a 
cubical  monument.  It  is  required  to  ascertain  how  many 
blocks,  or  feet,  will  be  the  length  of  one  side  of  the  monument. 

It  is  evident  that  the  number  of  blocks  will  be  equal  to  the 
cube  root  of  46,656.  As  the  given  number  consists  of  five  fig- 
ures, its  cube  root  will  contain  two  places  ;  for  the  cube  of  any 
number  can  never  contain  more  than  three  times  that  number, 
and  at  least  but  two  less.  We  therefore  separate  the  given 
number  into  periods  of  three  figures  each,  putting  a  point  over 
the  unit  figure,  and  every  third  figure  beyond  the  place  of 
units  ;  thus,  46,656.  We  find  by  the  table  of  powers,  or  by 
trial,  the  greatest  power  in  the  left-hand  period,  46  (thousand), 
is  27  (thousand),  the  root  of  which  is  3.  This  root  we  write  in 
the  quotient ;  and,  as  it  will  occupy  the  place  of  tens,  its  real 
value  is  30.  If  this  be  considered  the  side  of  a  cube,  it  will 
contain  27,000  cubic  feet,  30  x  30  X  30  =  27,000  feet. 


Fig.  1. 


Let  this  cube  be  represented  by  fig- 
ure 1,  each  of  whose  sides  measures 
30  feet ;  therefore  its  contents  will  be 
30  x  30  x  30  —  27,000  feet,  as  above. 
We  subtract  the  contents  of  this  cube 
from  46,656,  and  there  remain  19656 
cubic  feet. 


OPERATION. 


46,656(36  Ans. 
27,000 


2700)19,656 

16,200 

3,240 

216 

19^656 


Or  we  might  have  subtracted  the 
cube  of  3,  =  27,  from  the  first  period, 
and  to  the  remainder  have  brought 
down  the  next  period,  and  the  result 
would  have  been  the  same.  (See  op- 
eration.) The  cubic  blocks  that  re- 
main must  be  applied  to  the  three 
sides  of  figure  1.  For,  unless  a  cube 


250 


CUBE   ROOT. 


[SECT.  LXII. 


be  equally  increased  on  three  sides,  it  ceases  to  l>e  a  cube. 
To  effect  this,  we  must  find  the  superficial  contents  of  three 
sides  of  the  cube,  and  with  these  we  must  divide  the  remain- 
ing number  of  cubic  feet  or  blocks,  and  the  quotient  will  show 
the  thickness  of  the  additions.  As  the  length  of  a  side  is  30 
feet,  the  superficial  contents  will  be  30  X  30  =  900  square 
feet,  and  this  multiplied  by  3,  the  number  of  sides,  will  be 
900  X  3  =  2700  feet  With  this  as  a  divisor,  we  inquire  how 
many  times  it  is  contained  in  19,656,  and  find  it  to  be  6  times 
(one  or  two  units,  and  sometimes  three,  must  be  allowed  on 
account  of  the  other  deficiencies  in  enlarging  the  cube).  This 
6  is  the  thickness  of  the  additions  to  be  made  to  the  three 
sides  of  the  cube,  and  by  multiplying  their  superficial  contents 
by  it,  we  have  the  solid  contents  of  the  additions  to  be  made 
2700  X  6  =  16200  ;  that  is,  we  multiply  the  triple  square  by 
the  last  quotient  figure,  and  this  may  be  represented  by  the 
three  superficies  A  B  C  D,  E  F  G  H,  and  I  K  L  M.  (See 
figure  2.) 

Having  applied  these  additions  to  our  cube,  we  find  there 
are  three  other  deficiencies,  a  b  c  <2,  efg  h,  ij  k  /,  the  length 
of  which  is  equal  to  that  of  the  additions,  30  feet,  and  the 
height  and  breadth  of  each  are  equal  to  the  thickness  of  the  ad- 
ditions, 6  feet.  To  find  the  contents  of  these,  we  multiply  the 
product  of  their  length,  breadth,  and  thickness  by  their  num- 
ber ;  thus,  6x6x30x3  =  3240 ;  or,  which  is  the  same  thing, 
we  multiply  the  triple  quotient  by  the  square  of  the  last  quo- 
tient figure  ;  thus,  90  X  6  X  6  =  3240.  See  rule. 


Fig.  2; 


Having  made  these  additions  to 
the  cube,  we  still  find  one  other 
deficiency,  N.  (See  figure  2  )  The 
length,  breadth,  and  thickness  of 
which  are  equal  to  the  thickness  of 
the  former  additions,  viz.  6  feet. 
The  contents  of  this  are  found  by 
multiplying  its  length,  breadth,  and 
thickness  together ;  that  is,  cubing 
the  last  quotient  figure  ;  thus,  6x6 
X  6  =:  2 16.  By  making  this  last 
addition,  we  find  that  our  cubical 
monument  is  finished,  and  that  the 
first  figure  together  with  the  several  additions  is  equal  to  the  cu- 
bical blocks,  46,656. 


'a                          b/    / 

J*/ 

)                        C 

e      J 
S 

BA 


SECT.  LXH.]  CUBE  ROOT.  251 

Proof. 

27000  =  contents  of  fig.  1. 
16200=      "       "      first  additions. 
3240  =      "       "      second  additions. 
216  =      "       "      third  addition. 


46656  =  contents  of  the  whole  monument. 

I.  Required  the  cube  root  of  77308776. 

OPERATION. 

77308776(426  root.  4  X  4  X  300  —  4800 

64  4x    30—    120 

4920)  13308  =  1st  dividend.  1st  divisor  =  4920 

9600  4800x2  =  9600 

480  120  x  2  x  2  =    480 

8  2x2x2= 8 

10088  =  1st  subtrahend.  1st  subtrahend  =  10088 

530460)1*220776  =  2d  dividend.     42  x  42  x  300  =  529200 

3175200  42  x  30  =  1260 

45360  2d  divisor  =  530460 

216  529200  x  6  =  3175200 

3220776  =  2d  subtrahend.  1260  x  6  X  6  =      45360 

6x6x6= 216 

2d  subtrahend  =  3220776 

2.  What  is  the  cube  root  of  34965783  ?  Ans.  327. 

3.  What  is  the  cube  root  of  436036824287  ?  Ans.  7583. 

4.  What  is  the  cube  root  of  84.604519  ?          Ans.  4.39. 

5.  Required  the  cube  root  of  54439939.  Ans.  379. 

6.  Extract  the  cube  root  of  60236288.  Ans.  392. 

7.  Extract  the  cube  root  of  109215352.  Ans.  478. 

8.  What  is  the  cube  root  of  116.930169  ?         Ans.  4.89. 

9.  What  is  the  cube  root  of  .726572699  ?         Ans.  .899. 
10.  Required  the  cube  root  of  2.  Ans.  1. 25994-. 

II.  Find  the  cube  root  of  11.  Ans.  2.2239+. 

12.  What  is  the  cube  root  of  122615327232  ?  Ans.  4968. 

13.  What  is  the  cube  root  of  £f $  ?  Ans.  f. 

14.  What  is  the  cube  root  of  £ff  £  ?  Ans.  tf. 

15.  What  is  the  cube  root  of  ££f  ?  Ans.  f . 

16.  What  is  the  cube  root  of  ||f  £f  ?  Ans-  H- 
To  find  the  cube  root  of  any  number  mentally,  less  than 

1 ,000,000,  when  the  number  has  an  exact  root. 


252  CUBE   ROOT.  [SECT   LXII. 

RULE.  —  As  there  will  be  two  figures  in  the  root,  the  first  may  easily 
be  found  mentally,  or  by  the  table  of  powers  ;  and  if  t)ie  unit  figure  of 
the  power  be  1,  the  unit  figure  in  the  root  will  be  1  ;  and  if  it  be  8,  the 
root  will  be  2  ;  and  if  7  it  will  be  3  ;  and  if  the  unit  of  the  power  be  6, 
the  unit  of  the  root  will  be  6  ;  and  if  5,  it  will  be  5  ;  if  3,  it  will  be  7  ; 
if  2,  it  will  be  8  ;  and  if  the  unit  of  the  power  be  9,  the  unit  of  the  root 
will  be  9.  This  will  appear  evident  by  inspecting  the  table  of  powers. 

17.  What  is  the  cube  root  of  97336  ?  Ans.  46. 
Explanation.     By  examining  the  left-hand  period,  we  find 

the  root  of  97  is  4,  and  the  cube  of  4  is  64.  The  root  cannot 
be  5,  because  the  cube  of  5  is  125.  The  unit  of  the  power  is 
6  ;  therefore,  by  the  above  rule,  the  unit  figure  in  the  root  is  6. 
The  answer,  therefore,  is  46. 

18.  What  is  the  cube  root  of  132651  ?  Ans.  51. 

19.  What  is  the  cube  root  of  148877  ?  Ans. 

20.  What  is  the  cube  root  of  175616  ?  Ans. 

21.  What  is  the  cube  root  of  185193  ?  Ans. 

22.  What  is  the  cube  root  of  238328  ?  Ans. 

23.  What  is  the  cube  root  of  262144  ?  Ans. 

24.  What  is  the  cube  root  of  389017  ?  Ans. 

25.  What  is  the  cube  root  of  405224  ?  Ans. 

26.  What  is  the  cube  root  of  531441  ?  Ans. 

27.  What  is  the  cube  root  of  24389  ?  Ans. 

28.  What  is  the  cube  root  of  42875  ?  Ans. 


SECOND  METHOD  OF  EXTRACTING  THE  CUBE  ROOT. 

The  following  rule  for  the  extraction  of  the  root  of  the  third 
power,  though  it  is  essentially  the  same  with  the  former,  may 
yet  serve  to  make  the  reasons  for  the  several  steps  of  the  op- 
eration more  intelligible  to  the  learner. 

RULE.  —  Separate  tJie  number  whose  root  is  to  be  found  into  periods, 
as  under  the  former  rule,  and  find  by  trial  the  greatest  root  in  the  left- 
hand  period,  and  put  it  in  the  place  of  the  quotient. 

Subtract  the  third  power  of  this  root  from  the  period  to  which  it  be- 
longs, and  to  the  remainder  bring  down  the  next  period  for  a  dividend. 

Then,  to  find  a  divisor,  annex  a  cipher  to  the  root  already  found,  and 
multiply  twice  the  number  thus  formed  by  the  number  itself,  and  to  the 
product  add  the  second  power  of  this  number.* 

*  This  is  the  same  as  multiplying  the  square  of  the  radical  figure  by 
300,  as  in  the  former  rule. 


SECT.  MIL]  CUBE  ROOT.  253 

Ascertain  how  many  limes  this  divisor  is  contained  in  the  dividend, 
and  write  the  result  in  the  quotient.* 

T/ien,  to  find  the  subtrahend,  multiply  this  divisor  by  its  quotient,  and 
write  the  product  under  the  dividend.  To  this  add  three  times  the  pre- 
ceding radical  figure  ivith  a  cipher  annexed,^  multiplied  by  the  second 
power  of  the  figure  last  obtained,  and  also  the  third  power  of  this  last 
figure.  Subtract  the  sum  of  their  several  products  from  the  dividend 
above  them,  and  to  the  remainder  bring  down  the  next  period  for  a  new 
dividend.  With  the  parts  of  the  root  already  found  proceed  to  find  a 
divisor  and  subtract  as  above,  and  so  on,  till  the  successive  figures  of 
the  root  are  all  obtained. 

The  rationale  of  the  above  rule  may  be  made  to  appear  by 
the  solution  of  the  following  question. 

Let  it  be  required  to  find  the  cube  root  of  17576. 

OPERATION. 

20  X  2  =  40 
20 

800  17^76  (20  +  6  =  26  Ans. 

20  X  20  =    400  8000 

Divisor       1200  )  9576  dividend. 

7200 
2160 
216 

~9576  subtrahend. 

We  now  raise  the  quantity  20  +  6  to  the  third  power. 
20  +  6 
20  +  6 
400  + 120 

120  +  36 

400  +  240  +  36 
20+6 


8000  +  4800  +  720 

2400+1440  +  216 

8000  +  7200  +  2160  +  216  =  17576. 


*  This  quotient  figure  must  sometimes  be  less  than  the  one  indicated 
by  the  divisor,  and  in  extreme  cases  the  divisor  may  give  a  quotient  too 
large  by  several  units.  The  quotient  required  can,  of  course,  never  ex- 
ceed 9. 

t  This  accounts  for -multiplying  by  30,  in  the  foregoing  rule,  which  is  a 
factor  in  the  triple  quotient  in  finding  the  subtrahend. 
22 


254  CUBE  ROOT.  [SECT.  LXII. 

Now,  by  observing  this  operation,  and  remarking  what 
would  be  lost  in  the  course  of  it  by  omitting  the  second  figure 
of  the  root,  6,  taking  20  instead  of  26,  we  see  that  when  we 
have  found  the  20  the  next  inquiry  is,  what  number  must  be 
added  to  20,  so  that,  if  we  multiply  it  into  itself  once  and  into  20 
twice,  and  the  sum  of  these  products,  together  with  the  second 
power  of  twenty,  by  twenty  plus  this  number,  the  result  will  be 
17576,  or  8000  +  9576.  Then,  in  order  to  obtain  this  num- 
ber, or  an  approximation  to  it,  we  take  twice  the  part  of  the 
root  found,  20,  and  multiply  the  result,  40,  by  20,  as  we  should 
do  in  raising  it  to  the  third  power,  and  make  this,  which  is 
800,  a  part  of  the  divisor,  and  the  product  of  which  by  6  was 
lost  in  the  operation  for  want  of  the  6  added  to  20.  But  this  is 
not  all  the  loss.  There  is  also  the  second  power  of  20  by  6, 
and  therefore  400  to  be  added  to  the  800  for  a  divisor.  There 
still  remains  the  further  loss  of  the  third  power  of  6  (216),  and 
also  of  6  times  240  and  20  times  36  ;  but  these  we  neglect  in 
the  formation  of  the  divisor.  The  divisor  is  contained  in  the 
dividend  7  times ;  but,  making  the  allowance  of  a  unit  for  the 
neglect  of  the  numbers  above  named,  we  take  6  for  the  quo- 
tient figure,  and  proceed  to  find  the  subtrahend,  which,  accord- 
ing to  the  rule  and  the  foregoing  operation  of  raising  20  -|-  6  to 
the  third  power,  must  be  1200x6+1440  +  720  +  216=: 
17576. 


APPLICATION  OF  THE  CUBE  ROOT. 

PRINCIPLES  ASSUMED. 

Spheres  are  to  each  other  as  the  cubes  of  their  diameters. 
Cubes,  and  all  similar  solid  bodies,  are  to  each  other  as  the 
cubes  of  their  diameters,  or  homologous  sides. 

29.  If  a  ball,  3  inches  in  diameter,  weigh  4  pounds,  what 
will  be  the  weight  of  a  ball  that  is  6  inches  in  diameter  ? 

Ans.  321bs. 

30.  If  a  globe  of  gold,  one  inch  in  diameter,  be  worth  $120, 
what  is  the  value  of  a  globe  3A  inches  in  diameter  ? 

Ans.  $  5145. 

31.  If  the  weight  of  a  well-proportioned   man,  5  feet  10 
inches   in  height,  be  180  pounds,  what  must  have  been  the 
weight  of  Goliath  of  Gath,  who  was  10  feet  4f  inches  in  height  ? 

Ans.  1015.1+lbs. 

32.  If  a  bell,  4  inches  in  height,  3  inches  in  width,  and  £  of 


SECT.  LXII.]  GENERAL  RULE  FOR  ROOTS.  255 

an  inch  in  thickness-,  weigh  2  pounds,  what  should  be  the  dimen- 
sions of  a  bell  that  would  weigh  2000  pounds  ? 

Ans.  3ft.  4in.  high,  2ft.  Gin.  wide,  and  2£in.  thick. 

33.  Having  a  small  stack  of  hay,  5  feet  in  height,  weighing 
Icwt.,  I  wish  to  know  the  weight  of  a  similar  stack  that  is  20 
feet  in  height.  Ans.  64cwt. 

34.  If  a  man  dig  a  small  square  cellar,  which  will  measure 
6  feet  each  way,  in  one  day,  how  long  would  it  take  turn  to 
dig  a  similar  one  that  measured  10  feet  each  way  ? 

Ans.  4.629+  days. 

35.  If  an  ox,  whose  girth  is  6  feet,  weighs  GOOlbs.,  what  is 
the  weight  of  an  ox  whose  girth  is  8  feet  ?   Ans.  1422.2+lbs. 

36.  Four  women  own  a  ball  of  butter,  5  inches  in  diameter. 
It  is  agreed  that  each  shall  take  her  share  separately  from  the 
surface  of  the  ball.      How  many  inches  of  its  diameter  shall 
each  take  ? 

Ans.  First,  .45+^ inches  ;  second,  .62-}-  inches  ;  third, 
.78-}- inches ;  fourth,  3.15-)- inches. 

37.  John  Jones  has  a  stack  of  hay  in  the  form  of  a  pyramid. 
It  is  16  feet  in  height,  and  12  feet  wide  at  its  base.     It  contains 
5  tons  of  hay,  worth  $  17.50  per  ton.     Mr.  Jones  has  sold  this 
hay  to  Messrs.  Pierce,  Rowe,  Wells,  and  Northend.     As  the 
upper  part  of  the  stack  has  been  injured,  it  is  agreed  that  Mr. 
Pierce,  who  takes  the  upper  part,  shall  have  10  per  cent,  more 
of  the  hay  than  Mr.  Rowe  ;  and  Mr.  Rowe,  who  takes  his  share 
next,  shall  have  8  per  cent,  more  than  Mr.  Wells  ;  and  Mr. 
Northend,  who  has  the  bottom  of  the  stack,  that  has  been  much 
injured,  shall  have  10  per  cent,  more  than  Mr.  Wells.     Re- 
quired the  quantity  of  hay,  and  how  many  feet  of  the  height  of 
the  stack,  beginning  at  the  top,  each  receives. 

Ans.  Pierce  receives  275-5T47rcwt.  and  10.365-j-  feet  in  height ; 
Rowe,  24£ffcwt.  and  2.293+  feet;  Wells,  22ff$cwt.  and 
1.646+  feet ;  Northend,  25^cwt.  and  1.494+  feet. 


A  GENERAL  RULE  FOR  EXTRACTING  THE  ROOTS 
OF  ALL  POWERS. 

RULE.  —  1.  Prepare  the  given  number  for  extraction,  by  pointing  off 
from  the  unit's  place,  as  the  required  root  directs. 

2.  Find  the  first  figure  of  the  root  by  trial,  or  by  inspection,  in  the 
table  of  powers,  and  subtract  its  power  from  the  left-hand  period. 


256  GENERAL  RULE  FOR  ROOTS.  [SECT.  LXII. 

0 

3.  To  the  remainder,  bring  down  the  first  figure  in  the  next  period, 
and  call  it  the  dividend. 

4.  Involve  the  root  to  tJie  next  inferior  power  to  that  which  is  given, 
and  multiply  it  by  the  number  denoting  the  given  power  for  a  divisor. 

5.  Find  how  many  times  the  divisor  is  contained  in  the  dividend,  and 
the  quotient  will  be  another  figure  of  the  root.     An  allcnoance  of  two  or 
three  units  is  generally  made,  and  for  the  higher  powers  a  still  greater 
allowance  is  necessary. 

6.  Involve  the  whole  root  to  the  given  power,  and  subtract  it  from 
the  given  number,  as  before. 

7.  Bring  down  the  first  figure  of  the  next  period  to  the  remainder 
for  a  new  dividend,  to  which  find  a  new  divisor,  as  before ;  and  in  like 
manner  proceed  till  the  whole  is  finished. 

1.  What  is  the  cube  root  of  20346417  ? 

OPERATION. 

20346417(273 

23  =  _8_  =  l*t  subtrahend. 

22  X  3  =  12)  123  =  1st  dividend. 

273  —  19683          =  2d  subtrahend. 

2T2  X  3  =  2187)  6634       =  2d  dividend 
2733=  20346417    =  3d  subtrahend. 

2.  What  is  the  fourth  root  of  34828517376  ? 

OPERATION". 

34828517376(432.  Ans. 

44  =  256  =  1st  subtrahend. 

43  X  4  =       256)  922  =  1st  dividend. 

434  =  3418801  =  2d  subtrahend. 

433  X  4  =  318028)  640507         =  2d  dividend. 
432<  =  34828517376  =  3d  subtrahend. 

3.  What  is  the  5th  root  of  281950621875  ?          Ans.  195. 

4.  Required  the  sixth  root  of  1178420166015625. 

Ans.  325. 

5.  Required  the  seventh  root  of  1283918464548864. 

Ans.  144. 

6.  Required  the  eighth  root  of  218340105584896. 

Ans.  62. 


SECT.  LXIII.]        ARITHMETICAL  PROGRESSION.  257 

SECTION  LXIII. 
ARITHMETICAL   PROGRESSION. 

WHEN  a  series  of  quantities  or  numbers  increases  or  de- 
creases by  a  constant  difference,  it  is  called  arithmetical  pro- 
gression, or  progression  by  difference.     The  constant  difference 
is  called  the  common  difference  or  the  ratio  of  the  progression. 
Thus,  let  there  be  the  two  following  series :  — 

1,     5,     9,    13,  17,  21,  25,  29,  33, 
25,  22,  19,  16,  13,  10,    7,     4,     1. 

The  first  is  called  an  ascending  series  or  progression.     The 
second  is  called  a  descending  series  or  progression. 

The  numbers  which  form  the  series  are  called  the  terms  of 
the  progression. 

The  first  and  last  terms  of  the  progression  are  called  the 
extremes,  and  the  other  terms,  the  means. 

Any  three  of  the  five  following  things  being  given,  the  other 
two  may  be  found  :  — 

1st,  the  first  term, 

2d,  the  last  term, 

3d,  the  number  of  terms, 

4th,  the  common  difference, 

5th,  the  sum  of  the  terms. 

PROBLEM  I. 

The  first  term,  last  term,  and  number  of  terms  being  given, 
to  find  the  common  difference. 

To  illustrate  this  problem,  we  will  examine  the  following 
series,  — 

3,  5,  7,  9,  11,  13,  15,  17,  19. 

It  will  be  perceived  that  in  this  series  3  and  19  are  the  ex- 
tremes, 2  the  common  difference,  9  the  number  of  terms,  and 
99  the  sum  of  the  series. 

It  is  evident,  that  the  number  of  common  differences  in  any 
number  of  terms  will  be  one  less  than  the  number  of  terms. 
Hence,  if  there  be  9  terms,  the  number  of  common  differences 
will  be  8,  and  the  sum  of  these  common  differences  will  be 
equal  to  the  difference  of  the  extremes ;  therefore  if  the  differ- 
ence of  the  extremes,  19  —  3  —  16,  be  divided  by  the  number 
of  common  differences,  the  quotient  will  be  the  common  differ- 
ence. Thus  16  -r-  8  =  2  is  the  common  difference. 
22* 


258  ARITHMETICAL  PROGRESSION.        [SECT.  LXTII. 

RULE.  —  Divide  the  difference  of  the  extremes  by  the  number  of 
terms  less  one,  and  the  quotient  is  the  common  difference. 

1.  The  extremes  are  3  and  45,  and  the  number  of  terms  is 
22.     What  is  the  common  difference  ? 

OPERATION. 

45  —  3 

^-^  =  2  Answer. 

2.  A  man  is  to  travel  from  Albany  to  a  certain  place  in  11 
days,  and  to  go  but  5  miles  the  first  day,  increasing  the  dis- 
tance equally  each  day,  so  that  the  last  day's  journey  may  be 
45  miles.      Required  the  daily  increase.  Ans.  4  miles. 

3.  A  man  had  10  sons,  whose  several  ages  differed  alike  ; 
the  youngest  was  3  years,  and  the  oldest  48.     What  was  the 
common  difference  of  their  ages  ?  Ans.  5  years. 

4.  A  certain  school  consists  of  19  scholars ;  the  youngest  is 
3  years  old,  and  the  oldest  39.     What  is  the  common  differ- 
ence of  their  ages  ?  Ans.  2  years. 

PROBLEM  II. 

The  first  term,  last  term,  and  number  of  terms  being  given, 
to  find  the  sum  of  all  the  terms. 

Illustration. 

Let    3,    5,    7,    9,  11, 13, 15, 17,  19,  be  the  series, 
and  19,  17,  15,  13,  11,    9,    7,    5,    3,  the  same  series  inverted. 

22,  22,  22,  22,  22, 22, 22,  22, 22,  sum  of  both  series. 

From  the  arrangement  of  the  above  series,  we  see  that,  by 
adding  the  two  as  they  stand,  we  have  the  same  number  for 
the  sum  of  the  successive  terms,  and  that  the  sum  of  both 
series  is  double  the  sum  of  either  series. 

It  is  evident  that  if,  in  the  above  series,  22  be  multiplied  by 
9,  the  number  of  terms,  the  product  will  be  the  sum  of  both 
series,  22  X  9  =  198,  and  therefore  the  sum  of  either  series 
will  be  198  -r-  2  =  99.  But  22  is  also  the  sum  of  the  ex- 
tremes in  either  series,  3  -f-  19  =  22.  Therefore,  if  the  sum 
of  the  extremes  be  multiplied  by  the  number  of  terms,  the  prod- 
uct will  be  double  the  sum  of  the  series. 

RULE.  —  Multiply  the  sum  of  the  extremes  by  the  number  of  terms, 
and  half  the  product  will  be  the  sum  of  tlie  series. 

5.  The  extremes  of  an  arithmetical  series  are  3  and  45,  and 
the  number  of  terms  22.  Required  the  sum  of  the  series. 


SECT.  LXHI.]         ARITHMETICAL  PROGRESSION.  259 

OPERATION. 

45  +  3  X  22 

=  528  Answer. 

6.  A  man  going  a  journey  travelled  the  first  day  7  miles, 
the  last  day  51  miles,  and  he  continued  his  journey  12  days. 
How  far  did  he  travel  ?  Ans.  348  miles. 

7.  In  a  certain  school  there  are  19  scholars  ;  the  youngest  is 
3  years  old,  and  the  oldest  39.    What  is  the  sum  of  their  ages  ? 

Ans.  399  years. 

8.  Suppose  a  number  of  stones  were  laid  a  rod  distant  from 
each  other,  for  thirty  miles,  and  the  first  stone  a  rod  from  a 
basket.     What  length  of  ground  will  that  man  travel  over  who 
gathers  them  up  singly,  returning  with  them  one  by  one  to  the 
basket  ?  Ans.  288090  miles  2  rods. 

PROBLEM   III. 

The  extremes  and  the  common  difference  being  given,  to 
find  the  number  of  terms. 

Illustration.  —  Let  the  extremes  be  3  and  19,  and  the  common 
difference  2.  The  difference  of  the  extremes  will  be  19  —  3 
==  16  ;  and  it  is  evident,  that,  if  the  difference  of  the  extremes 
be  divided  by  the  common  difference,  the  quotient  is  the  num- 
ber of  common  differences  ;  thus  16  -=-  2  =  8.  We  have  de- 
monstrated in  Problem  I.  that  the  number  of  terms  is  one 
more  than  the  number  of  differences ;  therefore  8  -f-  1  =  9, 
the  number  of  terms. 

RULE.  —  Divide  the  difference  of  the  extremes  by  the  common  differ- 
ence, and  the  quotient  increased  by  one  toill  be  the  number  of  terms  re- 
quired. 

9.  If  the  extremes  are  3  and  45,  and  the  common  difference 
2,  what  is  the  number  of  terms  ? 

OPERATION. 
45  —  3 


-[-  1  —  22  Answer. 


10.  In  a  certain  school  the  ages  of  all  the  scholars  differ 
alike  ;  the  oldest  is  39  years,  the  youngest  is  3  years,  and  the 
difference  between  the  ages  of  each  is  2  years.      Required  the 
number  of  scholars.  Ans.  19. 

11.  A  man  going  a  journey  travelled  the  first  day  7  miles, 
the  last  day  51  miles,  and  each  day  increased  his  journey  by  4 
miles.     How  many  days  did  he  travel  ?  Ans.  12. 


260  ARITHMETICAL  PROGRESSION.        [SECT.  LXIII 

PROBLEM  IV. 

The  extremes  and  common  difference  being  given,  to  find 
the  sum  of  the  series. 

Illustration.  —  Let  the  extremes  be  3  and  19,  and  the  com- 
mon difference  2.  The  difference  of  the  extremes  will  be  19 

3—  16  ;  and  it  has  been  shown  in  the  last  problem,  that,  if 

the  difference  of  the  extremes  be  divided  by  the  common  dif- 
ference, the  quotient  will  be  the  number  of  terms  less  one  ; 
therefore  the  number  of  terms  less  one  will  be  16  -5-  2  =  8, 
and  the  number  of  terms  8  -f-  1  =  9.  It  was  demonstrated  in 
Problem  II.  that  if  the  number  of  terms  was  multiplied  by  the 
sum  of  the  extremes,  and  the  product  divided  by  2,  the  quo- 
tient would  be  the  sum  of  the  series. 

RULE.  —  Divide  the  difference  of  the  extremes  by  the  common  differ- 
ence, and  add  I  to  the  quotient;  multiply  this  quotient  by  the  sum  of  the 
extremes,  and  half  the  product  is  the  sum  of  the  series. 

12.  If  the  extremes  are  3  and  45,  and  the  common  differ- 
ence 2,  what  is  the  sum  of  the  series  ?  Ans.  528. 

13.  A  owes  B  a  certain  sum,  to  be  discharged  in  a  year,  by 
paying  6  cents  the  first  week,  18  cents  the  second  week,  and 
thus  to  increase  every  week  by  12  cents,  till  the  last  payment 
should  be  $  6. 18.     What  is  the  debt  ?  Ans.  $  162.24. 

PROBLEM  V. 

The  extremes  and  sum  of  the  series  being  given,  to  find  the 
common  difference. 

Illustration.  —  Let  the  extremes  be  3  and  19,  and  the  sum  of 
the  series  99,  to  find  the  common  difference.  We  have  before 
shown,  that,  if  the  extremes  be  multiplied  by  the  number  of 
terms,  the  product  would  be  twice  the  sum  of  the  series  ;  there- 
fore, if  twice  the  sum  of  the  series  be  divided  by  the  extremes, 
the  quotient  will  be  the  number  of  terms.  Thus,  99  X  2  — 
198  ;  3  -f-  19  =  22  ;  198  -~  22  =  9  is  the  number  of  terms. 
And  we  have  before  shown,  that,  if  the  difference  of  the  ex- 
tremes be  divided  by  the  number  of  terms  less  one,  the  quo- 
tient will  be  the  common  difference  ;  therefore,  19  —  3  =  16  ; 
9  —  1  zz:  8 ;  16-T-8  —  2is  the  common  difference. 

RULE.  —  Divide  twice  the  sum  of  tJte  series  by  the  sum  of  the  ex- 
tremes, and  from  the  quotient  subtract  1 ;  and  with  this  remainder  divide 
the  difference  of  the  extremes,  and  the  quotient  is  the  common  difference. 

14.  The  extremes  are  3  and  45,  and  the  sum  of  the  series 
528.     What  is  the  common  difference  ?  Ans.  2. 


BECT.  Liiv.j  GEOMETRICAL  SERIES.  261 

PROBLEM  VI. 

The  first  term,  number  of  terms,  and  the  sum  of  the  series 
being  given,  to  find  the  last  term. 

Illustration.  —  Let  3  be  the  first  term,  9  the  number  of 
terms,  and  99  the  sum  of  the  series.  By  Problem  II.  it  was 
shown,  that,  if  the  sum  of  the  extremes  were  multiplied  by  the 
number  of  terms,  the  product  was  twice  the  sum  of  the  series ; 
therefore,  if  twice  the  sum  of  the  series  be  divided  by  the  num- 
ber of  terms,  the  quotient  is  the  sum  of  the  extremes.  If  from 
this  we  subtract  the  first  term,  the  remainder  is  the  last  term  ; 
thus  99  X  2  =  198 ;  198  -7-  9  =  22  ;  22  —  3  =  19,  last  term. 

RULE.  —  Divide  twice  the  sum  of  the  series  by  the  number  of  terms  ; 
from  the  quotient  take  the  first  term,  and  the  remainder  will  be  the  last 
term. 

15.  A  merchant  being  indebted  to  22  creditors  $  528,  ordered 
his  clerk  to  pay  the  first  $  3,  and  the  rest  increasing  in  arith- 
metical progression.  What  is  the  difference  of  the  payments, 
and  the  last  payment  ? 

Ans.  Difference  2  ;  last  payment,  $  45. 


SECTION  LXIV. 
GEOMETRICAL    SERIES, 

OR  SERIES  BY  QUOTIENT. 

IF  there  be  three  or  more  numbers,  and  if  there  be  the  same 
quotient  when  the  second  is  divided  by  the  first,  and  the  third 
divided  by  the  second,  and  the  fourth  divided  by  the  third,  &c., 
those  numbers  are  in  geometrical  progression.  If  the  series 
increase,  the  quotient  is  more  than  unity ;  if  it  decrease,  it  is  less 
than  unity. 

The  following  series  are  examples  of  this  kind  :  — 
2,  6,  18,  54,  162,  486. 
64,  32,  16,  8,  4,  2. 

The  former  is  called  an  ascending  series,  and  the  latter  a 
descending  series. 

In  the  first,  the  quotient  is  3,  and  is  called  the  ratio ;  in  the 
second,  it  is  J. 


262  GEOMETRICAL  SERIES.  [SECT.  LIIT. 

The  first  and  last  terms  of  a  series  are  called  extremes,  and 
the  other  terms  means. 

PROBLEM  I. 

One  of  the  extremes,  the  ratio,  and  the  number  of  terms 
being  given,  to  find  the  other  extreme. 

Let  the  first  term  be  3,  the  ratio  2,  and  the  number  of  terms 
8,  to  find  the  last  term. 

Illustration.  —  It  is  evident,  that,  if  we  multiply  the  first 
term  by  the  ratio,  the  product  will  be  the  second  term  ;  and,  if 
we  multiply  the  second  term  by  the  ratio,  the  product  will  be 
the  third  term ;  -and,  in  this  manner,  we  may  carry  the  series  to 
any  desirable  extent.  By  examining  the  following  series,  we 
find  that  3,  carried  to  the  8th  term,  is  384  ;  thus, 

(1.)  (2.)    (3.)    (4.)     (5.)     (6.)       (7.)        (S.) 

3,  6,  12,  24,  48,  96,  192,  384,  ascending  series. 
But  the  factors  of  384  are  3,  2,  2,  2,  2,  2,  2,  and  2  ;  there- 
fore the  continued  product  of  these  numbers  will  produce  384. 
But,  by  multiplying  2  by  itself  six  times,  and  that  product  by  3, 
is  the  same  as  raising  the  ratio,  2,  to  the  seventh  power,  and 
then  multiplying  that  power  by  the  first  term.  Hence  the  fol- 
lowing 

RULE.  —  Raise  the  ratio  to  a  power  whose  index  is  equal  to  the  num- 
ber of  terms  less  one ;  then  multiply  this  power  by  the  first  term,  and  the 
product  is  the  last  term,  or  other  extreme. 

Illustration.  —  In  the  above  question  the  number  of  terms  is 
8 ;  we  therefore  raise  2,  the  ratio,  to  the  seventh  power,  it 
being  one  less  than  8  ;  thus,  2x2x2x2x2x2  X  2  =  128. 
We  then  multiply  this  number  by  3,  the  first  term,  and  the 
'product  is  the  last  term  ;  thus,  128  x  3  =  384,  last  term. 

The  above  rule  will  apply  in  a  descending  series.     Let  the 
following  numbers  be  a  geometrical  descending  series  :  — 
384,  192,  96,  48,  24,  12,  6,  3,  descending  series. 

Let  the  first  term  be  384,  the  number  of  terms  8,  and  the 
ratio  £,  to  find  the  other  extreme.  By  the  above  rule,  we  raise 
the  ratio,  £,  to  the  seventh  power,  it  being  one  less  than  the 
number  of  terms,  8.  Thus,  £X£X£X£X£X£X£= 
T!F-  We  then  multiply  this  power  by  the  first  term ;  thus, 
T£S-  X  -f-4-  =  iff  =  3,  the  extreme  required. 

But  as  the  last  term,  or  any  term  near  the  last,  is  very 
tedious  to  be  found  by  continual  multiplication,  it  will  often  be 
necessary,  in  order  to  ascertain  it,  to  have  a  series  of  numbers 


SECT.  LXIV.]  GEOMETRICAL   SERIES.  263 

in  arithmetical  proportion,  called  indices  or  exponents,  begin- 
ning either  with  a  cipher  or  a  unit,  whose  common  difference 
is  one.  When  the  first  term  of  the  series  and  ratio  are  equal, 
the  indices  must  begin  with  a  unit,  and  in  this  case  the  prod- 
uct of  any  two  terms  is  equal  to  that  term  signified  by  the 
sum  of  their  indices. 

rpi        (  1,  2,  3,    4,    5,    6,  &c.,  indices  or  arithmetical  series. 
u  '  J  2,  4,  8,  16,  32,  64,  &c.,  geometrical  series. 

Now  6  +  6  =  12  =  the  index  of  the  twelfth  term,  and  64  X 
64  =  4096  =  the  twelfth  term. 

But  when  the  first  term  of  the  series  and  the  ratio  are  differ- 
ent, the  indices  must  begin  with  a  cipher,  and  the  sum  of  the 
indices  made  choice  of  must  be  one  less  than  the  number  of 
terms  given  in  the  question ;  because  1  in  the  indices  stands 
over  the  second  term,  and  2  in  the  indices  over  the  third  term, 
&c.  And,  in  this  case,  the  product  of  any  two  terms,  divided 
by  theirs/,  is  equal  to  that  term  beyond  the  first  signified  by 
the  sum  of  their  indices. 

T,        ( 0,  1,  2,  3,     4,     5,      6,     &c.,  indices. 

'  {  1,  3,  9,  27,  81,  243,  729,  &c.,  geometrical  series. 

Here,  6  -f  5  =  11,  the  index  of  the  12th  term. 

729  X  243=  177147,  the  12th  term,  because  the  first  term 
of  the  series  and  ratio  are  different,  by  which  means  a  cipher 
stands  over  the  first  term. 

Thus,  by  the  help  of  these  indices,  and  a  few  of  the  first 
terms  in  any  geometrical  series,  any  term  whose  distance  from 
the  first  term  is  assigned,  though  it  were  ever  so  remote,  may 
be  obtained  without  producing  all  the  terms. 

1.  If  the  first  term  be  4,  the  ratio  4,  and  the  number  of 
terms  9,  what  is  the  last  term  ? 

OPERATION. 

1.  2.    3.      4   +    4    =     8 

4.  16.  64.  256  X  256  =  65536  =  power  of  the  ratio,  whose 
exponent  is  less  by  1  than  the  number  of  terms.  65536  X  4, 
the  first  term  =  262144  ==  last  term. 


Or,  4  X  4  =  262144  =  last  term,  as  before. 

ae  262144,  the  ratio  J,  E 
last  term? 

x  au^  —  ^i#  =4,  the  last  term. 


2.  If  the  first  term  be  262144,  the  ratio  J,  and  the  number 
of  terms  9,  what  is  the  last  term  ?  Ans.  4. 


364  GEOMETRICAL   SERIES.  [SECT.  LXIV. 

3.  If  the  first  term  be  72,  the  ratio  £,  and  the  number  of 
terms  6,  what  is  the  last  term  ?  Ans.  /y. 

4.  If  I  were  to  buy  30  oxen,  giving  2  cents  for  the  first  ox, 
4  cents  for  the  second,  8  cents  for  the  third,  &c.,  what  would 
be  the  price  of  the  last  ox  ?  Ans.  $10737418.24. 

5.  If  the  first  term  be  5,  and  the  ratio  3,  what  is  the  seventh 
term  ?  Ans.  3645. 

6.  If  the  first  term  be  50,  the  ratio  1.06,  and  the  number  of 
terms  5,  what  is  the  last  term  ?  Ans.  63.123848. 

7.  What  is  the  amount  of  $160.00  at  compound  interest  for 
6  years  ?  Ans.  $  226.96,305796096. 

8.  What  is  the  amount  of  $  300.00  at  compound  interest  at  5 
per  cent,  for  8  years  ?  Ans.  $  443.23,6+. 

9.  What  is  the  amount  of  8100.00  at  6  per  cent,  for  30 
years  ? 

Ans.  $  574.34,9117291325011626410633231080264584635- 
7252196069357387776. 

PROBLEM  II. 

The  first  term,  the  ratio,  and  the  number  of  terms  being 
given,  to  find  the  sum  of  all  the  terms. 

In  order  that  the  pupil  may  understand  the  following  rule, 
we  will  examine  a  question  analytically. 

Let  the  following  be  a  geometrical  series,  and  we  wish  to 
obtain  its  sum  :  — 

1,     3,     9,  27,  81. 

Illustration.  —  By  examining  this  series,  we  find  the  first 
term  to  be  1,  the  last  term  81,  the  ratio  3.  If  we  multiply  each 
term  of  the  following  series,  1,  3,  9,  27,  81,  by  3,  the  ratio, 
their  product  will  be  3,  9,  27,  81,  243,  and  the  sum  of  this  last 
series  will  be  three  times  as  much  as  the  first  series.  The  dif- 
ference, therefore,  between  these  series  will  be  twice  as  much 
as  ihejirst  series. 

3,     9,  27,  81,  243  =  second  series. 

1,     3,     9,  27,  81,          =  first  series. 

~Q,  0,  0,  0,  243  —  1  =  242,  difference  of  the  series. 
As  this  difference  must  be  twice  the  sum  of  the  first  series, 
therefore  the  sum  of  the  first  series  must  be  242  -4-  2  =  121. 

By  examining  the  above  series,  we  find  the  terms  in  both 
the  same,  with  the  exception  of  the  first  term  in  the  first  series, 
and  the  last  term  in  the  second  series.  We  have  only,  then,  to 


SECT.  LXIV.]  GEOMETRICAL  SERIES.  265 

subtract  the  first  term  in  the  first  series  from  the  last  term  in 
the  second  series,  and  the  remainder  is  twice  the  sum  of  the 
first  series  ;  and  half  of  this  being  taken  gives  the  sum  of  the 
series  required. 

RULE.  —  Find  the  other  extreme,  as  before,  multiply  it  by  the  ratio, 
and  from  the  product  subtract  the  given  extreme.  Divide  the  remainder 
by  the  ratio  less  1  (unless  the  ratio  be  less  than  a  unit,  in  which  case 
the  ratio  must  be  subtracted  from  1  )  ,  and  the  quotient  multiplied  by 
the  other  extreme  will  give  the  sum  of  the  series.  See  operation,  ques- 
tion 10.  Or,  raise  the  ratio  to  a  power  whose  index  is  equal  to  the  num- 
ber of  terms  ;  from  which  subtract  1  ,  divide  the  remainder  by  the 
ratio  less  1,  and  tJue  quotient,  multiplied  by  the  given  extreme,  will 
give  the  sum  of  the  series.  See  operation,  question  11. 

But  if  the  ratio  be  a  fraction  less  than  a  unit,  raise  the  ratio  to  a 
power  whose  index  shall  be  equal  to  the  number  of  terms  ;  subtract  this 
power  from  1  ,  divide  the  remainder  by  the  difference  between  1  and  the 
ratio,  and  the  quotient,  multiplied  by  the  given  extreme,  will  give  the 
sum  of  the  series  required.  See  operation,  question  12. 

10.  If  the  first  term  be  10,  the  ratio  3,  and  the  number  of 
terms  7,  what  is  the  sum  of  the  series  ?  Ans.  10930. 

OPERATION. 

3X3X3X3X3X3X10  =  7290,  last  term.  _ 
7290X3  =  21870;  21870  —  10  =  21860;  21860-*-  3—1 
=  10930  Ans. 

11.  If  the  first  term  be  4,  the  ratio  3,  and  the  number  of 
terms  5,  what  is  the  sum  of  the  series  ?  Ans.  484. 

OPERATION. 

3X  3X3X3X3=  243;  243  —  1  =  242; 

242  -f-  3—  1  =  121  ;  121  X  4  =  484  Ans. 

12.  If  the  first  term  be  6,  the  ratio  f  ,  and  the  number  of 
terms  4,  what  is  the  sum  of  the  series  ?  Ans. 

OPERATION. 

f  X  I  X  f  X  f  =  i&65  ;  1  -  f  f  I  ;  f  f  I  -  *V65  - 


X  *  =  ¥^  =  9^  Ans. 

13.  How  large  a  debt  may  be  discharged  in  a  year,  by  pay- 
ing  $  1  the  first  month,  $  10  the  second,  and  so  on,  in  a  tenfold 
proportion,  each  month  ?  Ans.  $111111111111. 

14.  A  gentleman  offered  a  house  for  sale,  on  the  following 
terms  ;  that  for  the  first  door  he  should  charge   10  cents,  for 
the  second  20  cents,  for  the  third  40  cents,  and  so  on  in  a  geo- 

23 


266  GEOMETRICAL   SERIES.  [SECT.  LXIY 

metrical  ratio,  there  being  40  doors.     What  was  the  price  of 
the  house  ?  Ans.  $  109951  162777.50. 

15.  If  the  first  term  be  50,  the  ratio  1.06,  and  the  number  of 
terms  4,  what  is  the  sum  of  the  series  ?  Ans.  218.7308. 

16.  A  gentleman  deposited  annually  $10  in  a  bank,  from  the 
time  his  son  was  born  until  he  was  20  years  of  age.     Required 
the  amount  of  the  deposits  at  6  per  cent.,  compound  interest, 
when  his  son  was  21  years  old.  Ans.  $  423.92,2+. 

17.  If  the  first  term  be  7,  the  ratio  J,  and  the  number  of 
terms  5,  what  is  the  sum  of  the  series  ?  Ans.  928/F. 

18.  If  one  mill  had  been  put  at  interest  at  the  commence- 
ment of  the  Christian  era,  what  would  it  amount  to  at  com- 
pound interest,  supposing  the  principal  to  have  doubled  itself 
every  12  years,  January  1,  1837  ? 

Ans.  $  114179815416476790484662877555959610910619- 
72.99,2. 

If  this  sum  was  all  in  dollars,  it  would  take  the  present  inhab- 
itants of  the  globe  more  than  1,000,000  years  to  count  it.  If  it 
was  reduced  to  its  value  in  pure  gold,  and  was  formed  into  a 
globe,  it  would  be  many  million  times  larger  than  all  the  bodies 
that  compose  the  solar  system. 

PROBLEM  III. 

To  find  the  sum  of  the  second  powers  of  any  number  of 
terms,  whose  roots  differ  by  unity. 

RULE.  —  Add  one  to  the  number  of  terms,  and  multiply  this  sum  by 
the  number  of  terms  ;  then  add  one  to  twice  live  number  of  terms,  and 
multiply  this  sum  by  the  former  product,  and  the  last  product,  divided 
by  6,  will  give  the  sum  of  all  the  terms. 

19.  What  is  the  sum  of  10  terms  of  the  series  I2,  22,  32,  42, 
52,  62,  72,  82,  92,  102  ? 

OPERATION. 


IPX 

-  Lg—  -  =  385  Ans. 

20.  What  is  the  sum  of  100  terms  of  the  series  I2,  22,  S2, 
42,  52,  62,  72,  82,  92,  102,  &c.,  to  1002  ?  Ans.  338350. 

21.  Purchased  50  lots  of  land  ;  the  first  was  one  rod  square, 
the  second  was   two  rods  square,  the   third  was   three   rods 
square,  and  so  on,  the  last  being  50  rods  square.     How  many 
square  rods  were  there  in  the  50  lots  ?  Ans.  42925. 

22.  Let  it  be  required  to  find  the  number  of  cannon  shot  in 
a  square  pile,  whose  side  is  80.  Ans.  173880.     . 


SECT.  LXV.]  INFINITE  SERIES.  267 

NOTE.  —  A  square  pile  is  formed  by  continued  horizontal  courses  of 
shot  laid  one  above  another,  and  these  courses  are  squares,  whose  sides 
decrease  by  unity  from  the  bottom  of  the  pile  to  the  top  row,  which  is 
composed  of  only  one  shot. 

PROBLEM  IV. 

To  find  the  sum  of  the  third  power  of  any  number  of  terms, 
whose  roots  differ  by  unity. 

RULE.  —  Add  one  to  the  number  of  terms,  and  multiply  this  sum  by 
half  the  number  of  terms;  the  square  of  this  product  is  the  sum  of  all 
the  series. 

23.  Required  the  sum  of  the  following  series  :  —  I3,  23,  33, 
43,  53,  63,  73,  83,  93,  103,  II3,  123. 

OPERATION. 

12  +  1=13;  12-f-2  =  6;  13X6  =  78;  78 X 78  =  6084 Ans. 

24.  I  have  10  blocks  of  marble,  each  of  which  is  an  exact 
cube.     A  side  of  the  first  cube  measures  one  foot,  a  side  of  the 
second  2  feet,  a  side  of  the  third  3  feet,  and  so  on  to  the  10th, 
whose  side  measures  10  feet.     Required  the  number  of  cubical 
feet  in  the  blocks  ?  Ans.  3025  cubic  feet. 

25.  What  is  the  sum  of  50  terms  of  the  series  I3,  23,  33,  43, 
53,  63,  73,  &c.,  up  to  503  ?  Ans.  1625625. 


SECTION  LXV. 
INFINITE  SERIES. 

AN  INFINITE  SERIES  is  such  as,  being  continued,  would  run 
on  ad  infinilum  ;  but  the  nature  of  its  progression  is  such,  that, 
by  having  a  few  of  its  terms  given,  the  others  to  any  extent 
may  be  known.  Such  are  the  following  series  :  — 

1,     2,     4,     8,     16,     32,     64,     128,  &c.,  ad  infnitum. 

125,  25,  5,     1,      £,     2-V»  T£¥,     ^,  &c.,  ad  infinitum. 


To  find  the  sum  of  a  decreasing  series. 

RULE.  —  Multiply  the  first  term  by  the  ratio,  and  divide  the  product 
by  the  ratio  less  1,  and  the  quotient  is  the  sum  of  an  infinite  decreasing 
series. 

1.  What  is  the  sum  of  the  series  4,  1,  £,  T^,  ^V>  &c.,  con- 
tinued to  an  infinite  number  of  terms  ? 

OPERATION. 

4  v  4 


=  5    Answer. 


268  DISCOUNT  BY  COMPOUND  INTEREST.  [SECT.  LXVI. 

2.  What  is  the  sum  of  the  series  5,  1,  •£,  ^,  dec.,  continued 
to  infinity  ?  Ans.  6£. 

3.  If  the  following  series,  8,  f ,  T%,  -yf  7,  &c.,  were  carried  to 
infinity,  what  would  be  its  sum  ?  Ans.  9£. 

4.  What  is  the  sum  of  the  following  series,  carried  to  in- 
finity :   1,  £,  £,  2V,  -g\,  &c.  ?  Ans.  1£. 

5.  What  is  the  sum  of  the  following  series,  carried  to  infin- 
ity:  ll,-YSH>&c.?  Ans.  12  f. 

6.  If  the  series  f ,  £,  J,  -j^-,  2^,  &c.,  were  carried  to  infinity, 
what  would  be  its  sum  ?  Ans.  1.    ^ 


SECTION  LXVI. 
DISCOUNT    BY    COMPOUND    INTEREST. 

1.  What  is  the  present  worth  of  $  600.00,  due  3  years  hence, 
at  6  per  cent,  compound  interest  ? 

OPERATION. 

1.06)3=  1.191016)600.00($  503.77+  Ans. 

By  analysis.  —  We  find  the  amount  of  $  1   at  compound 
interest  for  3  years  to  be   $  1.191016;  therefore   $  1  is  the 
present  worth  of  $  1.191016  due  3  years  hence.     And  if  $  1  is 
the  present  worth  of  $  1.191016,  the  present  worth  of 
600 

=  $503.77,1+ 


RULE.  —  Divide  the  debt  by  the  amount  of  one  dollar  for  the  given 
time,  and  the  quotient  is  the  present  worthy  which,  if  subtracted  from  the 
debt,  will  leave  the  discount. 

2.  What  is  the  present  worth  of  $  500.00,  due  4  years  hence, 
at  6  per  cent,  compound  interest  ?  Ans.  $  396.04,6+. 

3.  What  is  the   present  worth  of  $1000.00,  due   10  years 
hence,  at  5  per  cent,  compound  interest  ?  Ans.  $  613.91,3-)-. 

4.  What  is  the  discount  on  $800.00,  due  2  years  hence,  at 
6  per  cent,  compound  interest  ?  Ans.  $  88.00,3-)-. 

5.  What  is  the  present  worth  of  $  1728,  due  5  years  hence, 
at  6  per  cent,  compound  interest  ?  Ans.  $  1353.93. 

6.  What  is  the  discount  on  $  3700,  due  10  years  hence,  at  5 
per  cent,  discount,  compound  interest?  Ans.  $2271.47. 

7.  What  is  the  present  worth  of  $  7000,  due  2  years  hence', 
at  5  per  cent,  compound  interest  ?  Ans.  $  63492.10. 


SECT.  LXVII.]   ANNUITIES  AT  COMPOUND  INTEREST.         269 

SECTION  LXVII. 
ANNUITIES   AT   COMPOUND    INTEREST. 

AN  annuity  is  a  certain  sum  of  money  to  be  paid  at  regular 
periods,  either  for  a  limited  time  or  for  ever. 

The  present  worth  or  value  of  an  annuity  is  that  sum  which, 
being  improved  at  compound  interest,  will  be  sufficient  to  pay 
the  annuity. 

The  amount  of  an  annuity  is  the  compound  interest  of  all  the 
payments  added  to  their  sum. 

To  find  the  amount  of  an  annuity  at  compound  interest. 

RULE.  —  Make  $  1.00  the  first  term  of  a  geometrical  series,  and  the 
amount  of  $1.00  at  the  given  rate  per  cent,  the  ratio.  Carry  the  series 
to  so  many  terms  as  the  number  of  years,  and  find  its  sum.  Multiply 
the  sum  thus  found  by  the  given  annuity,  and  the  product  will  be  the 
amount. 

EXAMPLES. 

1.  What  will  an  annuity  of  $60  per  annum,  payable  yearly, 
amount  to  in  4  years,  at  6  per  cent.  ? 

1  +  1.06  +  1.06+  1.06  =  4.374616. 
4.374616  X  60  =  $  262.47,6+  Answer. 

°f'     L<^~  \  X  60  =  $262.47,6+  Answer. 
1.06  —  1 

2.  What  will  an  annuity  of  $  500.00  amount  to  in  5  years,  at 
6  per  cent.  ?  Ans.  $  2818.54,6+. 

3.  What   will   an   annuity   of    $1000.00,    payable   yearly, 
amount  to  in  10  years  ?  Ans.  $13180.79,4+. 

4.  What  will  an  annuity  of  $  30.00,  payable  yearly,  amount 
to  in  3  years  ?  Ans.  $  95.50,8  +. 

To  find  the  present  worth  of  an  annuity. 
As  the  first  payment  is  made  at  the  end  of  the  year,  its  pres- 
ent worth  or  value  is  a  sum  that  will  amount  in  one  year  to 
that  payment ;  and  as  the  second  payment  is  made  at  the  end 
of  the  second  year,  its  value  is  a  sum  that  will,  at  compound 
interest,  amount  in  two  years  to  that  payment ;  and  the  same 
principle  is  adopted  for  the  third  year,  fourth  year,  &c.     This 
may  be  illustrated  in  the  following  question. 
23* 


270        ANNUITIES  AT  COMPOUND  INTEREST.  [SECT.  LXVJI. 

5.  What  is  the  present  worth  of  an  annuity  of  $1.00,  to  con- 
tinue 5  years,  at  compound  interest  ? 

The  present  worth  of  $  1.00  for  1  year  =  $  0.943396 
The  present  worth  of  $  1.00  for  2  years  =  $  0.889996 
The  present  worth  of  $  1.00  for  3  years  —  $  0.839619 
The  present  worth  of  $  1.00  for  4  years  =  $  0.792094 
The  present  worth  of  $  1.00  for  5  years  =  $  0.747258 

$1.212363 

By  the  above  illustration,  we  perceive  that  the  present  worth 
of  an  annuity  of  $  1,  to  continue  5  years,  is  $  4.2 1  £>-{-.  Hence, 
having  found  the  present  worth  of  an  annuity  of  $  1  for  any 
given  time  by  Section  LXVL,  the  present  worth  of  any  other 
sum  may  be  found  by  multiplying  it  by  the  present  worth  of  $1 
for  that  time. 

RULE.  —  Multiply  the  present  worth  of  the  annuity  of  one  dollar  for 
the  given  time  by  the  given  annuity,  and  the  product  is  the  present  worth 
required.  Or,  find  the  amount  of  the  annuity  by  tlie  last  rule,  and  then 
find  its  present  worth. 

6.  What  is  the  present  worth  of  an  annuity  of  $  60,  to  bo 
continued  4  years,  at  compound  interest  ? 

First  Method. 

The  present  worth  of  $  1.00  for  1  year  =  $  0.943396 
The  present  worth  of  $  1.00  for  2  years  =  $  0.889996 
The  present  worth  of  $  1.00  for  3  years  =  $  0.839619 
The  present  worth  of  $  1.00  for  4  years  =  $  0.792093 

8  3.465104 
$  3.465104  X  60  =  $  207.90,6+  Answer. 

4  Second  Method. 

-1=  $  4.374616  x  .792093  =  $  3.465102  x  60  = 
$  207.90,6-f  Ans. 

7.  A  gentleman  wishes  to  purchase  an  annuity,  which  shall 
afford  him,  at  6  per  cent,  compound  interest,  $  500  a  year  for 
ten  years.     What  sum  must  he  deposit  in  the  annuity  office 
to  produce  it  ?  Ans.  $  3680.04-}-. 

8.  What  is  the  present  worth  of  an  annuity  of  $  1000,  to 
continue  10  years?  Ans.  $7360.08. 

9.  What  is  the  present  worth  of  an  annuity  of  $  1728,  to 
continue  3  years?  Ans.  $4618.96. 


SECT.  LXVII.]     ANNUITIES  AT  COMPOUND  INTEREST.        271 


By  the  assistance  of  the  following  tables,  questions  in  annui- 
ties may  be  easily  performed. 

TABLE  I. 

Showing  the  amount  of  $  1  annuity  from  1  year  to  40. 


Years. 

5  per  cent. 

6  per  cent. 

1  Years. 

5  per  cent. 

6  per  cent. 

1 

1.0000UO 

1.000000 

21 

35.719252 

39.992727 

2 

2.050000 

2.060000 

22 

38.505214 

43.392290 

3 

3.152500 

3  183600 

23 

41.430475 

46.995828 

4 

4.310125 

4.374616 

24 

44.501999 

50.815577 

5 

5.525631 

5.637093 

25 

47.727099 

54.864512 

6 

6  801913 

6.975319 

26 

51  113454 

59.156:583 

7 

8.142008 

8.393838 

27 

54.669126 

63.705766 

8 

9.549109 

9.897468 

28 

58.402583 

68.528112 

9 

11.026564 

11.491316 

29 

62.322712 

73  639798 

10 

12.577893 

13  180795 

30 

66.438847 

79  058186 

11 

14.206787 

.  14.971643 

31 

70.760790 

84.801677 

12 

15.917127 

16.869941 

32 

75.298829 

90.889778 

13 

17.712983 

18.882138 

33 

80.063771 

97.343165 

14 

19.598632 

21.015066 

34 

85.066959 

104.183755 

15 

21578564 

23.275970 

35 

90.220307 

111.434780 

16 

23.657492 

25.672528 

36 

95.836323 

119.120867 

17 

25.840366 

28.212880 

37 

101.628139 

127.268119 

18 

28132385 

30  905653 

38 

107.709546 

135.904206 

19 

30  539004 

33.759992 

39 

114.095023 

145.058458 

20 

33.065954 

36.785591 

40 

120.799774 

154.761966 

TABLE  II. 

Showing  the  present  value  of  an  annuity  of  $  1  from  1  year  to  40. 


Years. 

5  per  cent. 

6  per  cent. 

Years. 

5  per  cent. 

6  per  cent. 

1 

0  952381 

0.943396 

21 

12.821153 

11764077 

2 

1.859410 

1.833393 

22 

13.163003 

12.041582 

3 

2.723248 

2.673012 

23 

13.488574 

12.303379 

4 

3  545950 

3.465106 

24 

13.798642 

12.550358 

5 

4.329477 

4.212364 

25 

14.093945 

12.783356 

6 

5075692 

4.917324 

26 

14  375185 

13.003166 

7 

5  786373 

5.582381 

27 

14.643034 

13  210534 

8 

6.463213 

6.209794 

28 

14.898127 

13.406164 

9 

7.107822 

6.801692 

29 

15.141074 

13.590721 

10 

7.721735 

7.360087 

30 

15.372451 

13.764831 

11 

8  306414 

7.886875 

31 

15.592810 

13.929086 

12 

8.863252 

8.388844 

32 

15.802677 

14.084043 

13 

9.393573 

8.852683 

33 

16.002549 

14.230230 

14 

9.898641 

9.294984 

34 

16.192904 

14.368141 

15 

10379658 

9.712249 

35 

1  16.374194 

14.498246 

16 

10.837770 

10.105895 

36 

16.546852 

14.620987 

17 

11.274066 

10  477260 

37 

16.711287 

14  736780 

18 

11.639587 

10.827603 

38 

16.867893 

14.846019 

19 

12.085321 

11.158116 

39 

17.017041 

14.949075 

20 

12.462216 

11.469921 

40 

17.159086 

15.046297 

272  ASSESSMENT  OF  TAXES.  [SECT.  LXVIII. 

10.  What  is  the  present  worth  of  an  annuity  of  $  200,  at  5 
per  cent,  compound  interest,  for  7  years  ?    Ans.  $1157.27-}-. 

1 1.  What  is  the  present  worth  of  an  annuity  of  $  300,  to  con- 
tinue 8  years,  at  6  per  cent,  compound  interest  ? 

Ans.  $  1862.93,8-f . 

12.  What  is  the  present  value  of*  an  annuity  of  $  100,  at  6 
per  cent,  for  9  years  ?  Ans.  $  680.16,9+. 

Questions  to  be  performed  by  the  preceding  tables. 

13.  What  will  an  annuity  of  $  30  amount  to  in  1 1  years,  at 
6  per  cent.  ? 

By  Table  I.,  the  amount  of  $  1  for  11  years  is  $14.971643 ; 
therefore,  $  14.971643  X  30  =  $  449.14,9+  Answer. 

14.  What  is  the  present  worth  of  an  annuity  of  $  80  for  30 
years,  at  5  per  cent.  ? 

By  Table  II.,  the  present   worth   of  $  1   for  30  years  is 
$15.372451,  therefore  $15.372451  X  80  =  $  1229.79,6+  Ans. 

15.  What  will  an  annuity  of  $800  amount  to  in  25  years,  at 

5  per  cent.  ?  Ans.  $  38181.67,9+. 

16.  What  will  an  annuity  of  $40  amount  to  in  30  years,  at 

6  per  cent.  ?  Ans.  $  3162.32,7+. 

17.  Required  the  present  worth  of  an  annuity  of  $  500,  to 
continue  40  years,  at  6  per  cent.  Ans.  8  7523.14,8+. 

18.  A  certain  parish  in  the  town  of  B.,  having  neglected  for 
6  years  to  pay  their  minister's  salary  of  $  700,  what  in  justice, 
provided  he  has  preached  the  truth,  should  he  receive  ? 

Ans.  $  4882.72,3. 


SECTION  LXVIII. 
ASSESSMENT   OF   TAXES. 

A  TAX  is  a  duty  laid  by  government,  for  public  purposes, 
on  the  property  of  the  inhabitants  of  a  town,  county,  or  State, 
and  also  on  the  polls  *  of  the  male  citizens  liable  by  law  to  as- 
sessment. 

A  tax  may  be  either  general  or  particular ;  that  is,  it  may 
affect  all  classes  indiscriminately,  or  only  one  or  more  classes. 

*  Poll  is  said  to  be  a  Saxon  word,  meaning  head.  In  the  constitution 
it  means  a  person ;  that  is,  a  person  who  is  liable  to  taxation. 


SECT.  LXTIII.]  ASSESSMENT  OF  TAXES.  273 

Taxes  may  be  either  direct  or  indirect ;  that  is,  they  may 
either  be  imposed  on  the  incomes  or  property  of  individuals,  or 
on  the  articles  on  which  these  incomes  or  property  are  ex- 
pended. 

The  method  of  assessing  town  taxes  is  not  precisely  the  same 
in  all  the  States,  yet  the  principle  is  virtually  the  same. 

In  some  of  the  States  the  poll  tax  is  more  than  in  others. 

The  following  is  the  law  regulating  taxation  in  Massachusetts 
(see  Revised  Statutes,  page  79) :  — 

"  The  assessors  shall  assess  upon  the  polls,  as  nearly  as  the 
same  can  be  conveniently  done,  one  sixth  part  of  the  whole 
sum  to  be  raised  ;  provided  the  whole  poll  tax  assessed  in  any 
one  year  upon  any  individual  for  town  and  county  purposes, 
except  highway  taxes,  shall  not  exceed  one  dollar  and  fifty 
cents ;  and  the  residue  of  said  whole  sum  to  be  raised  shall  be 
apportioned  upon  property  " ;  that  is,  on  the  real  and  personal 
estate  of  individuals  which  is  taxable. 

RULE  FOR  ASSESSING  TAXES. — First  take  an  inventory  of  all  the 
taxable  property,  real  and  personal,  in  the  town  or  county,  and  then  the 
number  of  polls  liable  to  taxation.  Multiply  the  sum  assessed  on  each 
poll  by  the  number  of  taxable  polls  in  the  town.  Subtract  this  amount 
from  the  sum  to  be  raised  by  the  town.  TJien  as  the  whole  valuation  of 
the  town  is  to  the  sum  to  be  raised,  after  having  deducted  the  amount  to 
be  paid  by  the  polls,  so  is  the  amount  of  each  man's  real  and  personal 
estate  to  his  tax.  Or,  if  the  sum  to  be  raised  on  property  be  divided  by 
the  valuation  of  t/ie  town,  the  quotient  will  be  the  sum  to  be  paid  on  each 
dollar  of  an  individual"1  s  real  or  personal  estate.  Multiply  each  man's 
property  by  this  sum,  and  the  product  will  be  the  amount  of  his  taxes. 

The  town  of  B.  is  to  be  taxed  $4109.  The  real  estate  of 
the  town  is  valued  at  $  493,000,  and  the  personal  property  at 
$  177,000.  There  are  506  polls,  each  of  which  is  taxed  $1.50. 
What  is  John  Smith's  tax,  whose  real  estate  is  valued  at  $  3700, 
and  his  personal  at  $  2300,  he  paying  for  6  polls  ?  And  what 
will  be  the  tax  on  $  1.00  ? 

OPERATION. 

$  1.50  X  506  —  $759,  amount  assessed  on  the  polls. 

$  493,000  -f-  $177 ,000m  $  670,000,  amount  of  taxable  property. 

$  4109  —  $  759  =  $  3350,  amount  to  be  assessed  on  property. 

$  670,000  :  $  3350  ::$!:$  .005,  to  be  assessed  on  each  dollar. 

$3700  X  .005=  $18.50,  tax  on  Smith's  real  estate. 

$  2300  X  .005  =  $  11.50,  tax  on  Smith's  personal  estate. 

$  1.50  X  6  =  $  9.00,  tax  on  6  polls. 

$  18.50  +  $  11.50  +  $  9.00  =  $  39.00,  amount  of  Smith's  tax. 


274 


ASSESSMENT  OF  TAXES.  [SECT.  MVIII. 


What  will  be  the  amount  of  taxation  on  each  of  the  following 
individuals  of  the  above  town,  their  taxable  property  being  as 
annexed  to  their  names  ? 


Persons. 

James  Dow, 
John  Brown, 
Samuel  Foster, 
James  Emerson, 
A.  C.  Hasseltine, 


Real  Estate. 

$4780 
7500 
1135 
8960 
7140 


Personal  Estate.      No.  of  Polls. 


$1720 
2120 
175 
5000 
3720 


9 

Form  of  a  tax-list  committed  to  the  collector,  containing  the 
answers  to  the  above  questions. 


Names. 

No.  of 

Polls. 

Poll  Tax. 

ITscT 

1.50 
10.50 
3.00 
0.00 
9.00 

Tax  on 
Real 

Estate. 

Tax  on 
Personal 
Estate. 

$8.60 
10.60 
.87£ 
25.00 
18.60 
11.50 

Total. 

$37.00 
49.60 
17.05 
72.80 
54.30 
39.00 

Time 
when 
paid. 

James  Dow, 
John  Brown, 
Samuel  Foster, 
James  Emerson, 
A.  C.  Hasseltine, 
John  Smith, 

3 

1 

7 
2 
0 
6 

$23.90 
37.50 
5.67£ 
44.80 
35.70 
18.50 

Having  found  the  amount  to  be  raised  on  the  dollar,  the  op- 
eration of  assessing  taxes  will  be  much  facilitated  by  the  use  of 
the  following 

TABLE. 


$    $ 

$      $ 

$       $ 

1  gives  0.005 

40  gives  0.20 

700  gives  3.50 

2  "  0.010 

50  "  0.25 

800  "   4.00 

3  "  0.015 

60  "  0.30 

900  "   4.50 

4  "  0.020 

70  "  0.35 

1000  "   5.00 

5  "  0.025 

80  "  0.40 

2000  «  10.00 

6  "   0.030 

90  "  0.45 

3000  «  15.00 

7  "   0.035 

100  "  0.50 

4000  "  20.00 

8  "  0.040 

200  "  1.00 

5000  "  25.00 

9  "  0.045 

300  "  1.50 

6000  "  30.00 

10  "  0.050 

400  "  2.00 

7000  "  35.00 

20  "   0.100 

500  "  2.50 

8000  "  40.00 

30  "   0.150 

600  "  3.00 

9000  "  45.00 

By  the  aid  of  the  above  table  the  amount  of  any  person's  tax 
may  be  found. 

Required  the  amount  of  James  Dow's  tax,  his  real  estate 
being  $  4780,  his  personal  $  1720,  and  he  paying  for  3  polls. 


SICT.  LXIX.]  ALLIGATION.  275 

1.  To  find  the  amount  on  his  real  estate. 

OPERATION. 

Dow's  tax  on  $  4000  is  $  20.00 
"  "  700  "        3.50 

"  "  80  "          .40  r 

, [estate. 

$  4780       $  23.90  Amount  of  Dow's  tax  on  real 

2.  To  find  the  amount  on  his  personal  estate. 

OPERATION. 

Dow's  tax  on   $  1000  is  $  5.00 
"  "  700  "      3.50 

"  "  20  "        .10  r 

[sonal  estate. 

$  1720       $  8.60  Amount  of  Dow's  tax  on  per- 
Tax  on  real  estate,         $  23.90 
Tax  on  personal  estate,       8.60 
Tax  on  3  polls,  4.50 

Dow's  whole  tax,  $  37.00 

NOTE.  —  It  will  be  necessary  to  construct  a  different  table,  although  on 
the  same  principle,  when  a  different  per  cent,  is  paid  on  the  dollar. 


SECTION  LXIX. 
ALLIGATION. 

ALLIGATION  teaches  how  to  compound  or  mix  together  sev- 
eral simples  of  different  qualities,  so  that  the  composition  may 
be  of  some  intermediate  quality  or  rate.  It  is  of  two  kinds,  Al- 
ligation Medial  and  Alligation  Alternate. 

ALLIGATION  MEDIAL. 

Alligation  Medial  teaches  how  to  find  the  mean  price  of  sev- 
eral articles  mixed,  the  quantity  and  value  of  each  being  given. 

RULE.  —  As  the  sum  of  the  quantities  to  be  mixed  is  to  their  value, 
so  is  any  part  of  the  composition  to  its  mean  price. 

EXAMPLES. 

1.  A  grocer  mixed  2cwt.  of  sugar  at  $  9.00  per  cwt.,  and 
Icwt.  at  $7.00  per  cwt.,  and  2cwt.  at  $  10.00  per  cwt.  ;  what 
is  the  value  of  Icwt.  of  this  mixture  ? 


276  ALLIGATION.  [SECT.  LXIX. 

2cwt.  at  8  9.00  =  $  18.00 

1       »        7.00=       7.00 

_2       "      10.00  =     20.00 

5      "  :          845.00::  Icwt.  :  89.00  Answer. 

2.  If  19  bushels  of  wheat  at  $  1.00  per  bushel  should  be 
mixed  with  40  bushels  of  rye   at  $0.66  per  bushel,  and  11 
bushels  of  barley  at  $  0.50  per  bushel,  what  would  a  bushel  of 
the  mixture  be  worth  ?  Ans.  $  0.72,7f  . 

3.  If  3  pounds  of  gold  of  22  carats  fine  be  mixed  with  3 
pounds  of  20  carats  fine,  what  is  the  fineness  of  the  mixture  ? 

Ans.  21  carats. 

4.  If  I  mix  20  pounds  of  tea  at  70  cents  per  pound  with  15 
pounds  at  60  cents  per  pound,  and  80  pounds  at  40  cents  per 
pound,  what  is  the  value  of  1  pound  of  this  mixture  ? 

Ans. 


NOTE.  —  If  an  ounce,  or  any  other  Quantity,  of  pure  gold  be  divided 
into  24  equal  parts,  these  parts  are  called  carats.  But  gold  is  often  mixed 
with  some  baser  metal,  which  is  called  the  alloy  ;  and  the  mixture  is  said 
to  be  so  many  carats  fine,  according  to  the  proportion  of  pure  gold  con- 
tained in  it  ;  thus,  if  22  carats  of  pure  gold  and  2  of  alloy  be  mixed  to- 
gether, it  is  said  to  be  22  carats  fine. 


ALLIGATION  ALTERNATE. 

This  rule  teaches  us  how,  from  the  prices  of  several  articles 
given,  to  find  how  much  of  each  must  be  mixed  to  bear  a  cer- 
tain price. 

CASE  I. 

RULE.  —  Place  the  prices  under  each  other,  in  the  order  of  their  value; 
connect  the  price  of  each  ingredient,  tvhich  is  less  in  value  than  the  in- 
tended compound,  with  one  which  is  of  greater  value  than  the  compound. 
Place  tJie  difference  between  the  price  and  that  of  each  simple,  opposite  to 
the  price  with  which  they  are  connected. 

EXAMPLES. 

5.  A  merchant  has  spices,  some  at  18  cents  a  pound,  some 
at  24  cents,  some  at  48  cents,  and  some  at  60  cents.  How 
much  of  each  sort  must  he  mix  that  he  may  sell  the  mixture  at 
40  cents  a  pound  ? 

Ihs.  cts. 

20  at  18  ^ 

Mean  rate  40  <  ^     '   '   16   "  48  (  Answers- 
22   "  60  ) 


SECT.   LXIX.] 


Mean  rate  40 


ALLIGATION. 


cts.  Ibs.         c 

18 — i         8  at  ] 
24— i — |  20  "   24 
48 — l     I  22  "   48 
«n 1 


60 


16 


Mean  rate  40 


cts.          Ibs.    Ibs. 

Iba. 
28 

28 
38 

cts. 
at  18 
"  24 
"  48 

15— 

zu  - 
20- 

22  J 

-  o 
-  8 
-  16 

24—, 
48  — 

i 

Answers. 


Explanation.  —  By  connecting  the  less  rate  with  the  greater, 
and  placing  the  differences  between  them  and  the  mean  rate  al- 
ternately, the  quantities  resulting  are  such,  that  there  is  precise- 
ly as  much  gained  by  one  quantity  as  is  lost  by  the  other,  and 
therefore  the  gain  and  loss  upon  the  whole  must  be  equal,  and 
the  compound  will  have  the  value  of  the  proposed  rate ;  the 
same  will  be  true  of  any  other  two  simples  managed  according 
to  the  rule.  In  like  manner,  let  the  number  of  simples  be  what 
they  may,  and  with  how  many  soever  every  one  is  linked,  since 
it  is  always  a  less  with  a  greater  than  the  mean  price,  there 
will  be  an  equal  balance  of  loss  and  gain  between  every  two, 
and  consequently  an  equal  balance  on  the  whole. 

It  is  obvious,  from  the  rule,  that  questions  of  this  sort  admit 
of  a  great  variety  of  answers  ;  for  having  found  one  answer, 
we  may  find  as  many  others  as  we  please  by  only  multiplying 
or  dividing  each  of  the  quantities  found  by  2,  3,  or  4,  &c., 
the  reason  of  which  is  evident ;  for  if  two  quantities  of  two  sim- 
ples make  a  balance  of  loss  and  gain  with  respect  to  the  mean 
price,  so  must  also  the  double  or  treble,  the  half  or  third  part, 
or  any  other  equimultiples  or  parts  of  these  quantities. 

6.  How  much  barley  at  50  cents  a  bushel,  and  rye  at  75 
cents,  and  wheat  at  $  1.00,  must  be  mixed,  that  the  composi- 
tion may  be  worth  80  cents  a  bushel  ? 

Ans.  20  bushels  of  rye,  20  of  barley,  and  35  of  wheat. 

7.  A  goldsmith  would  mix  gold  of  19  carats  fine  with  some 
of  15,  23,  and  24  carats  fine,  that  the  compound  may  be  20 
carats  fine.     What  quantity  of  each  must  he  take  ? 

Ans.  4oz.  of  15  carats,  3oz.  of  19,  loz.  of  23,  and  5oz.  of  24. 

8.  It  is  required  to  mix  several  sorts  of  wine  at  60  cents,  80 
cents,  and  $  1.20,  with  water,  that  the  mixture  may  be  worth 
75  cents  per  gallon ;  how  much  of  each  sort  must  be  taken  ? 

Ans.  45gals.  of  water,  ogals.  of  60  cents,  15gals.  of  80  cents, 
and75gals.  of  $1.20. 
24 


278  ALLIGATION.  [SECT.  LXII. 

CASE  II. 

When  one  of  the  ingredients  is  limited  to  a  certain  quantity. 

RULE.  —  Take  the  difference  between  each  price  and  the  mean  rate,  as 
before;  then  say,  as  the  difference  of  that  simple  whose  quantity  is  given 
is  to  the  rest  of  the  differences  severally,  so  is  the  quantity  given  to  the 
several  quantities  required. 

EXAMPLES. 

9.  How  much  wine  at  5s.,  at  5s.  6d.,  and  at  6s.  a  gallon,  must 
be  mixed  with  3  gallons  at  4s.  per  gallon,  so  that  the  mixture 
may  be  worth  5s.  4d.  per  gallon  ? 

8  +  2=10         Then  10  :  10  : :  3  :  3 
8  +  2  =  10  10  :  20  : :  3  :  6 

16  -f  4  =  20  10  :  20  : :  3  :  6 

16  -|-  4  =  20 
Ans.  3gals.  at  5s.,  6  at  5s.  6d.,  and  6  at  6s. 

10.  A  grocer  would  mix  teas  at  12s.,  10s.,  and  6s.  per  pound, 
with  20  pounds  at  4s.  per  pound  ;  how  much  of  each  sort  must 
he  take  to  make  the  composition  worth  8s.  per  pound  ? 

Ans.  201bs.  at  4s.,  lOlbs.  at  6s.,  lOlbs.at  10s.,and201bs.  at  12s. 

11.  How  much  port  wine  at  $  1.75  per  gallon,  and  temper- 
ance wine  at  $  1.25  per  gallon,  must  be  mixed  with  20  gal- 
lons of  water,  that  the  whole  may  be  sold  at  $  1 .00  per  gallon  ? 

Ans.  20gals.  port  wine,  and  20gals.  temperance  wine. 

12.  How  much  gold  of  15,  17,  and  22  carats  fine  must  be 
mixed  with  5  ounces  of  18  carats  fine,  so  that  the  composition 
may  be  20  carats  fine  ? 

Ans.  5oz.  of  15  carats,  5oz.  of  17,  and  25oz.  of  22. 

CASE  III." 

When  the  sum  and  quality  of  the  ingredients  are  given. 

RULE.  —  Find  an  ansiver  as  before,  by  linking ;  then  say,  as  the  sum 
of  the  quantities  or  differences,  thus  determined,  is  to  the  given  quantity, 
so  is  each  ingredient  found  by  linking  to  the  required  quantity  of  each. 

*  To  this  case  belongs  the  curious  fact  of  King  Hiero's  crown. 

Hiero,  King  of  Syracuse,  gave  orders  for  a  crown  to  be  made  of  pure 
gold  ;  but  suspecting  that  the  workmen  had  debased  it,  by  mixing  it  with 
silver  or  copper,  he  recommended  the  discovery  of  the  fraud  to  the  famous 
Archimedes,  and  desired  to  know  the  exact  quantity  of  alloy  in  the  crown. 

Archimedes,  in  order  to  detect  the  imposition,  procured  two  other  mass- 
es, the  one  of  pure  gold,  the  other  of  copper,  and  each  of  the  same  weight 
of  the  former  ;  and  by  putting  each  separately  into  a  vessel  full  of  water, 
the  quantity  of  water  expelled  by  them  determined  their  specific  gravi- 


SECT.  LSI.]  PERMUTATIONS  AND  COMBINATIONS.  279 

EXAMPLES. 

13.  How  many  gallons  of  water  must  be  mixed  with  wine  at 
$  1.50  per  gallon,  so  as  to  fill  a  vessel  containing  100  gallons, 
that  it  may  be  sold  at  $  1.20  per  gallon  ? 

gals.       gals.         gals.    gala. 

19ft /° 1     30     150:100::    30  :  20  water  )  A 

U  {  150— I   120     150  :  100  : :  120  :  80  wine   J  * 

150 

14.  A  merchant  has  sugar  at  8  cents,  10  cents,  12  cents,  and 
20  cents  per  pound  ;  with  these  he  would  fill  a  hogshead  that 
would  contain  200  pounds.     How  much  of  each  kind  must  he 
take,  that  he  may  sell  the  mixture  at  15  cents  per  pound  ? 

Ans.  33£lbs.  of  8,  10,  and  12  cts.,  and  lOOlbs.  of  20  cts. 


SECTION  LXX. 
PERMUTATIONS   AND   COMBINATIONS. 

THE  permutation  of  quantities  is  the  showing  how  many  dif- 
ferent ways  the  order  or  position  of  any  given  number  of  things 
may  be  changed. 

The  combination  of  quantities  is  the  showing  how  often  a 
less  number  of  things  can  be  taken  out  of  a  greater,  and  com- 
bined together,  without  considering  their  places  or  the  order 
they  stand  in. 

CASE  I. 

To  find  the  number  of  permutations  or  changes  that  can  be 
made  of  any  given  number  of  things,  all  different  from  each 
other. 

RULE.  —  Multiply  all  the  terms  of  the  natural  series  of  numbers,  from 

ties  ;  from  which,  and  their  given  weights,  the  exact  quantities  of  gold 
and  alloy  in  the  crown  may  be  determined. 

Suppose  the  weight  of  each  crown  to  be  10  pounds,  and  that  the  water 
expelled  by  the  copper  was  92  pounds,  by  the  gold  52  pounds,  and  by  the 
compound  crown  64  pounds  :  what  will  be  the  quantities  of  gold  and  al- 
loy in  the  crown  ? 

The  rates  of  the  simples  are  92  and  52,  and  of  the  compound  64 ;  there- 
fore, 

92 — j  12  of  copper,  40  :  10  :  :  12  :  31bs.  of  copper  \  An«,wpr 
62_J  28  of  ^       40  :  10  :  :  28  :  Tibs,  of  gold      5  Ans 


280  PERMUTATIONS  AND  COMBINATIONS.  [SECT.  LXX. 

1  up  to  the  given  number,  continually  together,  and  the  last  product  will 
be  the  answer  required. 

This  rule  may  be  illustrated  by  inquiring  how  many  different 
numbers  may  be  formed  from  the  figures  of  the  following 
number,  789,  making  use  of  the  three  figures  in  each  number. 

OPERATION. 

789,  798,  879,  897,  978,  987. 

It  will  be  perceived,  that  six  are  all  the  permutations  the 
above  number  will  admit  of.  By  adopting  the  rule,  we  find  the 
same  answer. 

1X2X3  =  6  Ans. 

1.  How  many  changes  may  be  rung  on  6  bells  ? 

OPERATION. 

1X2X3X4X5X6  =  720  changes.  Ans. 

2.  For  how  many  days  can  10  persons  be  placed  in  a  differ- 
ent position  at  dinner  ?  Ans.  3628800. 

3.  How  many  changes  may  be  rung  on  12  bells,  and  how 
long  would  they  be  in  ringing,  supposing  10  changes  to  be  rung 
in  one  minute,  and  the  year  to  consist  of  365  days,  5  hours,  and 
49  minutes  ?          Ans.  479001600,  and  91y.  26d.  22h.  41m. 

4.  How  many  changes  or  variations  will  the  letters  of  the 
alphabet  admit  of?      Ans.  403291461126605635584000000. 

CASE  II. 

Any  number  of  different  things  being  given,  to  find  how 
many  changes  can  be  made  out  of-  them,  by  taking  any  given 
number  of  quantities  at  a  time. 

RULE.  —  Take  a  series  of  numbers,  beginning  at  the  number  of  things 
given,  and  decreasing  by  I,  to  the  number  of  quantities  to  be  taken  at  a 
time ;  the  product  of  all  the  terms  will  be  the  answer  required. 

Illustration.  —  The  above  rule  may  be  illustrated  by  inquiring 
how  many  different  numbers  can  be  made  by  a  selection  of  any 
three  figures  from  the  following  number,  1234. 

OPERATION. 

123,  124,  132,  134,  142,  143,  213,  214,  231,  234,  241,  243, 
312,  314,  321,  324,  341,  342,  412,  413,  421,  423,  431,  432. 

By  examining  the  above,  it  will  be  perceived  that  there  are 
24  different  numbers  or  permutations  ;  and  this  number  may 
be  obtained  by  multiplying  the  number  of  things  given,  4,  by 
the  next  lower  number,  and  that  product  by  the  next  lower,  and 


SECT.  LXX.]  PERMUTATIONS  AND   COMBINATIONS.  281 

so  continuing  the  multiplications  as  many  times  as  there  are 
things  to  be  taken  at  a  time ;  and  the  things  to  be  taken  at  a 
time  are  3 ;  we  therefore  find  the  continued  product  of  the  first 
three  numbers,  beginning  at  4 ;  thus, 

4  X  3  X  2  =  24  Answer. 

NOTE.  —  The  questions  under  this  rule  refer  to  permutations  and  not 
combinations. 

5.  How  many  changes  can  be  rung  with  4  bells  out  of  8  ? 

OPERATION. 

8X7X6X5=  1680  changes,  Answer. 

6.  How  many  words  can  be  made  with  6  letters  out  of  26  of 
the  alphabet,  admitting  a  word  might  be  made  from  consonants  ? 

Ans.  165765600. 

CASE  III. 

To  find  the  number  of  combinations  of  any  given  number  of 
things,  all  different  one  from  another,  taking  any  given  number 
at  a  time,  without  reference  to  their  arrangement. 

RULE.  — Take  the  series  1,  2,  3,  4,  <3fC.,  up  to  the  number  to  be  taken 
at  a  time,  and  find  the  product  of  all  tJie  terms. 

Take  a  series  of  as  many  terms,  decreasing  by  1  from  the  given  num- 
ber out  of  which  the  election  is  to  be  made,  and  find  the  product  of  all 
the  terms. 

Divide  the  last  product  by  the  former,  and  the  quotient  will  be  tJie 
number  required. 

7.  How  many  combinations  can  there  be  of  any  three  let- 
ters of  the  alphabet  out  of  any  four  letters,  without  reference  to 
their  arrangement  or  permutations  ? 

Illustration.  —  By  examining  four  letters,  a,  b,  c,  d,  we  find 
they  will  admit  of  only  four  combinations.  Thus,  abc,  abd, 
acd,  bed.  And  this  number  can  be  ascertained  in  the  following 
manner  :  —  1X2X3=6;  4X3X2  =  24  ;  24  -s-  6  =  4, 
the  number  of  combinations. 

8.  How  many  combinations  can  be  made  of  7  letters  out  of 
10,  the  letters  all  being  different  ?  Ans.  120. 

OPERATION. 

1X2X3X4X5X6X7=:  5040 ;  7  being  the  number 
taken  at  a  time. 

10X9X8X7X6X5X4  =  604800  =  same  number 
from  10. 

604800  —  5040  =  120  Ans.   Number  of  combinations. 
24* 


282  LIFE   INSURANCE.  [SECT.  LXXI. 


CASE  IV. 

To  find  the  compositions  of  any  number  of  things,  in  an 
equal  number  of  sets,  the  things  being  all  different. 

RULE.  —  Multiply  the  number  of  things  in  every  set  continually  to- 
gethert  and  tJie  product  will  be  the  answer  required. 

9.  Suppose  there  are  4  companies,  in  each  of  which  there 
are  9  me.n  ;  it  is  required  to  find  how  many  ways  4  men  may  be 
chosen,  one  out  of  each  company.    9x9x9x9  =  6561  Ans. 

10.  There  are  4  companies,  in  one  of  which  there  are  6  men, 
in  another  8,  and  in  each  of  the  other  two  9  men.     What  are 
the  choices,  by  a  composition  of  4  men,  one  out  of  each  com- 
pany ?  Ans.  3888. 

11.  How  many  changes  are  there  in  throwing  5  dice  ? 

Ans.  7776. 


SECTION  LXXI. 
LIFE  INSURANCE. 

INSURANCE  on  life  is  a  contract,  which  stipulates  for  the  pay- 
ment of  a  certain  sum  of  money  on  the  death  of  an  individual, 
in  consideration  of  the  immediate  payment  of  a  specified  sum, 
or  more  frequently  of  an  annuity,  or  annual  premium,  to  be 
continued  during  the  existence  of  the  life  insured.  .  Contracts 
of  this  kind  are  of  immense  importance  to  society.  Every 
man  whose  income  depends  on  his  own  life  or  exertions,  and 
on  whom  others  are  dependent  for  support,  must  be  sensible  of 
the  advantages  of  arrangements  by  means  of  which,  at  a  small 
sacrifice  of  immediate  comfort,  he  is  enabled  effectually  to 
provide  against  the  casualties  of  life.  Though  nothing  can  be 
more  uncertain  than  the  continuance  of  an  individual  life,  yet 
nothing  is  more  invariable  than  the  duration  of  life  in  the  mass ; 
consequently,  the  exact  value  of  life  insurances  can  be  calcu- 
lated without  any  uncertainty  whatever,  and  a  man  by  effecting 
an  insurance  secures  to  his  family,  against  risk  of  accident,  the 
advantages  they  would  have  from  enjoying  his  exact  proportion 
of  the  average  duration  of  life.  Such  transactions  provide 
against  destitution,  and  tend  directly  to  the  accumulation  of 
capital.  No  wise  man  will  therefore  neglect  to  provide  against 
contingencies. 


SECT.  LXXI.] 


LIFE   INSURANCE. 


283 


The  following  table  from  Milne's  Treatise  on  the  Valuation 
of  Annuities  and  Assurances  (Vol.  II.  p.  565)  shows  the  "  ex- 
pectation of  life  "  at  every  age,  according  to  the  law  of  mor- 
tality at  Carlisle.' 

Expectation  of  Life. 


Age. 

Expectation. 

Age. 

Expectation. 

Age. 

Expectation. 

Age. 

Expectation. 

0 

36.72 

26 

37.14 

52 

19.68 

78 

6.12 

1 

44.68 

27 

36.41 

53 

18.97 

79 

5.80 

2 

47.55 

28 

35.69 

54 

18.28 

80 

5.51 

3 

49.82 

29 

35.00 

55 

17.58 

81 

5.21 

4 

50.76 

30 

34.34 

56 

16.89 

82 

4.93 

5 

51.25 

31 

33.68 

57 

16.21 

83 

4.65 

6 

51.17 

32 

33.03 

58 

15.55 

84 

4.39 

7 

50.80 

33 

32.36 

59 

14.92 

85 

4.12 

8 

50.24 

34 

31.68 

60 

14.34 

86 

3.90 

9 

45.57 

35 

31.00 

61 

13.82 

87 

3.71 

10 

48.82 

36 

30.32 

62 

13.31 

88 

3.59 

11 

48.04 

37 

29.64 

63 

12.81 

89 

3.47 

12 

47.27 

38 

28.96 

64 

12.30 

90 

3.28 

13 

46.51 

39 

28.28 

65 

11.79 

91 

3.26 

14 

45.75 

40 

27.61 

66 

11.27 

92 

3.37 

15 

45.00 

41 

26.97 

67 

10.75 

93 

3.48 

16 

44.27 

42 

26.34 

68 

10.23 

94 

3.53 

17 

43.57 

43 

25.71 

69 

9.70 

95 

3.52 

18 

42.87 

44 

25.09 

70 

9.19 

96 

3.46 

19 

42.17 

45 

24.46 

71 

.   8.65 

97 

3.28 

20 

41.46 

46 

23.82 

72 

8.16 

98 

3.07 

21 

40.75 

47 

23.17 

73 

7.72 

99 

2.77 

22 

40.04 

48 

22.50 

74 

7.33 

100 

2.28 

23 

39.31 

49 

21.81 

75 

7.01 

101 

1.79 

24 

38.59 

50 

21.11 

76 

6.69 

102 

1.30 

25 

37.86 

51 

20.39 

77 

6.40 

103 

0.83 

By  the  above  table  it  will  be  perceived,  that  tj*e  average 
age  to  which  100  persons  will  live  from  birth  will  be  38T7^ 
years,  and  that  100  persons  having  attained  the  age  of  35  years 
will  live  on  an  average  31  years  longer ;  and  the  average  time 
which  100  persons  will  continue  to  live,  after  the  age  of  100 
years,  is  2-^  years.  Taking  the  above  table  as  the  basis  of 
their  calculations,  various  life-insurance  companies  have  been 
formed,  whose  tables  or  rates  of  insurance  vary  according  as 
they  charge  a  greater  or  less  per  cent,  on  their  capital.  We 
have  selected  two  tables  from  those  used  by  the  numerous  cor- 
porations which  have  been  formed  in  the  United  States.  The 
first  is  the  one  adopted  by  the  Massachusetts  Hospital  Life  In- 
surance Company,  and  the  other  by  the  New  York  Life  Insur- 
ance Company. 


284 


LIFE  INSURANCE. 


[SECT.  LXXI. 


Age. 

Massachusetts. 

New  York. 

1  Year 

7  Years. 

10  Years. 

1  Year. 

7  Years.        For  Life. 

14 

.82 

.84 

.85 

.72 

.86 

.53 

15 

.83 

.85 

.86 

.77 

•      .88 

.56 

16 

.84 

.86 

.87 

.84 

.90 

.62 

17 

.85 

.87 

.88 

.86 

.91 

.65 

18  . 

.86 

.88 

.89 

.89 

.92 

.69 

19 

.87 

.89 

.90 

.90 

.94 

.73 

20 

.88 

.90 

.92 

.91 

.95 

.77 

21 

.89 

.91 

.93 

.92 

.97 

.82 

22 

.90 

.92 

.94 

.94 

.99 

.88 

23 

.91 

.94 

.96 

.97 

.03 

.93 

24 

.92 

.95 

.98 

.99 

.07 

.98 

25 

.93 

.97 

.99 

1.00 

.12 

2.04 

26 

.95 

.98 

.01 

1.07 

.17 

2.11 

27 

.96 

1.00 

.03 

1.12 

.23 

2.17 

28 

.98 

'1.02 

.05 

1.20 

.28 

2.24 

29 

.00 

1.04 

.07 

1.28 

.35 

2.31 

30 

.01 

.06 

.09 

1.31 

.36 

2.36 

31 

.03 

.08 

.11 

1.32 

.42 

2.43 

32 

.05 

.10 

.14 

1.33 

.46 

2.50 

33 

.07 

.13 

.16 

1.34 

.48 

257 

34 

.09 

.15 

.19 

1.35 

.50 

2.64 

35 

.12 

.18 

.22 

1.36 

.53 

2.75 

36 

.14 

.20 

.25 

1.39 

.57 

2.81 

37 

.16 

.23 

.29 

1.43 

.63 

2.90 

38 

.19 

.27 

.34 

1.48 

.70 

3.05 

39 

.22 

.31 

.39 

1.57 

.76 

3.11 

40 

.25 

.35 

.44 

1.69 

.83 

3.20 

41 

.28 

.40 

.50 

1.78 

.88 

3.31 

42 

.31 

.46 

.58 

1.85 

.89  . 

3.40 

43 

.35 

.53 

1.66 

1.89 

.92 

3.51 

44 

.40 

.61 

1.75 

1.90 

.94 

3.63 

45 

.47 

.70 

1.85 

1.91 

.96 

3.73 

46 

.54 

.80 

1.95 

1.92 

.98 

3.87 

47 

.62 

.90 

2.07 

1.93 

.99 

4.01 

48 

.71 

2.01 

2.20 

1.94 

2.02 

4.17 

49 

.81 

2.14 

2.34 

1.95 

2.04 

4.49 

50 

1.91 

2.27 

2.49 

1.96 

2.09 

4.60 

51 

2.03 

2.42 

2.65 

1.97 

2.20 

4.75 

52 

2.16 

2.58 

2.83 

2.02 

2.37 

4.90 

53 

2.29 

2.75 

3.03 

2.10 

2.59 

5.24 

54 

2.44 

2.94 

3.24 

2.18 

2.89 

5.49 

55 

2.60 

3.14 

3.47 

2.32 

3.21 

5.78 

56 

2.78 

3.36 

3.72 

2.47 

3.56 

6.05 

57 

2.96 

3.61 

400 

2.70 

4.20 

6.27 

58 

3.17 

3.87 

4.29 

3.14 

4.31 

6.50 

59 

3.39 

4.17 

4.62 

3.67 

4.63 

6.75 

60 

3.64 

4.48 

4.97 

4.35 

4.91 

7.00 

SECT.  LXXI.J  LIFE  INSURANCE.  285 

The  preceding  table  shows  at  what  rate  a  hundred  dollars 
may  be  insured  on  a  person's  life  for  one  year,  for  seven  years, 
ten  years,  and  for  life,  at  any  age  from  14  to  60  years.  Thus, 
a  man  who  is  42  years  old,  and  who  wishes  to  obtain  a  life 
insurance  in  Massachusetts  for  one  year,  for  $  100,  must  pay 
$1.31,  and  in  the  same  proportion  for  a  larger  sum.  And,  if  he 
would  obtain  a  life  insurance  in  New  York,  he  must  pay  annu- 
ally $  3.40  on  every  hundred  dollars  for  which  he  obtains  insur- 
ance. 

EXAMPLES. 

1.  What  premium  will  the  Massachusetts  Hospital  Life  Insur- 
ance Company  require  for  the  insurance  of  a  life  for  1  year 
for  $  1728,  the  person  being  30  years  of  age  ?      Ans.  $  17.45. 

OPERATION. 

$  1728  X  .0101  =  $  17.45  Ans. 

NOTE.  —  As  the  premium  is  $  1.01  on  $  100,  the  sum  insured  must  be 
multiplied  by  ffi  —  .0101. 

2.  What  amount  of  premium  must  James  Kimball  pay  at  the 
above  office  to  effect  an  insurance  on  his  life  for  7  years  for 
$  8000,  his  age  being  53  years  ?  Ans.  $  220. 

OPERATION. 

$  8000  x  .0275  =  $  220  annually,  Ans. 

3.  John  Smith,  60  years  of  age,  wishes  to  engage  in  a  very 
profitable  speculation,  and,  being  destitute  of  the  necessary  funds, 
he  effects  an  insurance  on  his  life  for  10  years,  for  $78,000,  at 
the  office  of  the  above  Company.     Required  the  amount  of  the 
annual  premium.  Ans.  $3736.20. 

4.  What  will  be  the  premium  per  annum  for  insuring  a  per- 
son's life,  who  is  15  years  old,  for  82000  for  7  years,  at  the 
New  York  Life  Insurance  Company  ?  Ans.  $  17.60. 

5.  A  gentleman  45  years  of  age,  being  bound  on  a  long  and 
dangerous  voyage,  and  wishing  to  secure  a  competence  for  his 
family,  obtains  an  insurance  for  life  at  the  above  office  in  New 
York,  for  $  12,000.     By  an  act  of  Providence,  he  dies  before 
the  end  of  the  third  year.     What  is  the  net  gain  to  his  family  ? 

Ans.  8  10,657.20. 

6.  John  Swan,  20  years  old,  effects  an  insurance  for  life  at 
New  York,  for  $10,000,  for  which  he  pays  an  annual  premium 

.of  ^J  per  cent.  If  the  Insurance  Company  loan  the  premium 
at  6  per  cent,  compound  interest,  and  Swan  should  die  at  the 
age  of  60  years,  who  will  gain  by  the  insurance  ? 

Ans.  Company  will  gain  $  17,382.86+. 
NOTE.  —  All  premiums  are  paid  annually  and  in  advance. 


286  POSITION.  [SECT.  LXXII. 

SECTION  LXXH. 
POSITION. 

POSITION  is  a  method  of  performing  such  questions  as  cannot 
be  resolved  by  the  common  direct  rules,  and  is  of  two  kinds, 
called  single  and  double. 

SINGLE   POSITION. 

SINGLE  POSITION  teaches  us  to  resolve  those  questions  whose 
results  are  proportional  to  their  suppositions. 

RULE.  —  Take  any  number  and  perform  the  same  operations  with  it 
as  are  desirable  to  be  performed  in  the  question.  Then  say,  as  the  result 
of  the  operation  is  to  the  position,  so  is  the  result  in  the  question  to  the 
number  required. 

EXAMPLES. 

1.  A  schoolmaster  being  asked  how  many  scholars  he  had, 
replied,  that  if  he  had  as  many  more  as  he  now  had,  and  half 
as  many  more,  he  should  have  200;  of  how  many  did  his 
school  consist  ? 

Suppose  he  had  60     As  150  :  200  : :  60 

Then,  as  many  more  60  60 

Half  as  many  more  _30          15Q  j^J  (  8Q  ^^  Ang> 

150  12000 

By  analysis.  —  By  having  as  many  more,  and  half  as  many 
more,  he  must  have  2£  times  the  original  number ;  therefore, 
by  dividing  200  by  2£,  we  obtain  the  answer,  80,  as  before. 

NOTE.  —  Having  performed  all  the  following  questions  by  position,  the 
student  should  then  perform  them  by  analysis. 

2.  A  person  after  spending  £  and  J  of  his  money  had  $  60 
left ;  what  had  he  at  first?  Ans.  $  144. 

3.  What  number  is  that,  which,  being  increased  by  £,  £,  and 
i  of  itself,  the  sum  shall  be  125  ?  Ans.  60. 

4.  A's  age  is  double  that  of  B,  and  B's  is  triple  that  of  C, 
and  the  sum  of  all  their  ages  is  140.     What  is  each  person's 
age  ?  Ans.  A's  84,  B's  42,  C's  14  years. 

5.  A  person  lent  a  sum  of  money  at  6  per  cent.,  and  at  the 
end  of  10  years  received  the  amount  $  560.     What  was  the 
sum  lent  ?  Ans.  $  350. 


SECT.  LXXII.]  POSITION.  287 

6.  Seven  eighths  of  a  certain  number  exceed  ^  by  81 ;  what 
is  the  number  ?  Ans.  120. 

7.  What  number  is  that  whose  f  exceed  |-  by  2-J~£  ? 

Ans.  87. 

DOUBLE   POSITION.* 

DOUBLE  POSITION  teaches  to  resolve  questions,  by  making 
two  suppositions  of  false  numbers. 

Those  questions  in  which  the  results  are  not  proportional  to 
their  positions  belong  to  this  rule. 

RULE.  —  Take  any  two  convenient  numbers,  and  proceed  with  each  ac- 
cording to  the  conditions  of  the  question.  Find  how  much  the  results 
are  different  from  the  result  in  the  question.  Multiply  each  of  the  errors 
by  the  contra  supposition,  and  find  the  sum  and  difference  of  the  prod- 
ucts. If  the  errors  are  alike,  divide  the  difference  of  the  products  by  the 
difference  of  the  errors,  and  the  quotient  will  be  the  answer.  If  the  er- 
rors are  unlike,  divide  the  sum  of  the  products  by  the  sum  of  the  errors, 
and  the  quotient  will  be  the  answer.  . 

NOTE.  —  The  errors  are  said  to  be  alike  when  they  are  both  too  great, 
or  both  too  small ;  and  unlike  when  one  is  too  great  and  the  other  too  little. 

EXAMPLES. 

1.  A  lady  purchased  a  piece  of  silk  for  a  gown  at  80  cents 
per  yard,  and  lining  for  it  at  30  cents  per  yard  ;  the  gown  and 
lining  contained  15  yards,  and  the  price  of  the  whole  was 
$  7.00.  How  many  yards  were  there  of  each  ? 

Suppose  6  yards  of  silk,  value  $  4.80 

She  must  then  have  9  yards  of  lining,  value  2.70 

Sum  of  their  values,  $  7.50 

Which  should  have  been  7.00 

So  the  first  error  is  50  too  much,  -j-  .50 

Again  ;  suppose  she  had  4  yards  of  silk,  value  $  3.20 

Then  she  must  have  1 1  yards  of  lining,  value  3.30 

Sum  of  their  values,  $  6.50 

Which  should  have  been  7.00 

So  that  the  second  error  is  50  too  little,  — .50 

*  This  rule  is  founded  on  the  supposition,  that  the  first  error  is  to  the 
second  as  the  difference  between  the  true  and  first  supposed  number  is  to 
the  difference  between  the  true  and  second  supposed  number.  When 
this  is  not  the  case,  the  exact  answer  to  the  questions  cannot  be  found  by 
this  rule. 


288                                          POSITION.  [SECT.  LXIII. 

First  supposition  multiplied  by  last  error,  6  X  50  :=  3.00 

Last  supposition  multiplied  by  first  error,  4  X  50  =  2.00 

Add  the  products,  because  unlike^  $  5.00 


500  -f-  50  +50  =  5  yards  of  silk,  )  A  5  X  80  =  $  4.00 

15—  5  =  10  yards  of  lining,       f#  10  X  30  =     3.00 

Proof    $17.00 

By  Analysis.  —  As  the  silk  and  lining  contain  15  yards,  and 
cost  $7.00,  the  average  price  per  yard  is  46§  ;  and  this  taken 
from  80  leaves  33£  ;  and  30  taken  from  46f  leaves  16f  ;  and 
as  the  quantity  of  lining  will  be  to  that  of  the  silk  as  33£  to 
16f  ,  it  is  therefore  evident  that  the  quantity  of  lining  is  twice 
the  quantity  of  silk.  Wherefore,  if  15,  the  number  of  yards,  be 
divided  into  three  parts,  two  of  those  parts  (10)  will  be  the 
number  of  yards  for  the  lining,  and  the  other  part  (5)  will  be 
the  yards  for  the  silk,  as  before. 

NOTE.  —  The  student  should  perform  each  question  by  analysis. 

2.  A  and  B  invested  equal  sums  in  trade  ;  A  gained  a  sum 
equal  to  £  of  his  stock,  and  B  lost  $  225  ;  then  A's  money  was 
double  that  of  B's.     What  did  each  invest  ?          Ans.  $  600. 

3.  A  person  being  asked  the  age  of  each  of  his  sons,  replied, 
that  his  eldest  son  was  4  years  older  than  the  second,  his  second 
4  years  older  than  the  third,  his  third  4  years  older  than  the 
fourth,  or  youngest,  and  his  youngest  half  the  age  of  the  oldest. 
What  was  the  age  of  each  of  his  sons  ? 

Ans.  12,  16,  20,  and  24  years. 

4.  A  gentleman  has  two  horses  and  a  saddle  worth  $  50. 
Now,  if  the  saddle   be  put  on  the  first  horse,  it  will  make  his 
value  double  that  of  the  second  horse  ;  but  if  it  be  put  on  the 
second,  it  will  make  his  value  triple  that  of  the  first.     What  was 
the  value  of  each  horse  ?      Ans.  The  first  $  30,  second  $  40. 

5.  A  gentleman  was  asked  the  time  of  day,  and  replied,  that 
f  of  the  time  past  from  noon  was  equal  to  ^  of  the  time  to 
midnight.     What  was  the  time  ?         Ans.  12  minutes  past  3. 

6.  A  and  B  have  the  same  income.     A  saves  TV  of  his,  but 
B,  by  spending  $100  per  annum  more  than  A,  at  the  end  of  10 
years  finds  himself  $  600  in  debt.     What  was  their  income  ? 

Ans.  $  480. 

7.  A  gentleman  hired  a  laborer  for  90  days  on  these  condi- 
tions :   that  for  every  day  he  wrought  he  should  receive  60 
cents,  and  for  every  day  he  was  absent  he  should   forfeit  80 


SECT.  LXXII.]  POSITION.  289 

cents.     At  the  expiration  of  the  term  he  received  $  33.     How- 
many  days  did  he  work,  and  how  many  days  was  he  idle  ? 

Ans.  He  labored  75  days,  and  was  idle  15  days. 

The  following  question,  with  some  variation  in  the  language, 
is  taken  from  Fenn's  Algebra,  page  62.  It  is  believed,  howev- 
er, that  Sir  Isaac  Newton  was  the  author  of  it. 

8.  If  12  oxen  eat  3£  acfes  of  grass  in  4  weeks,  and  21  oxen 
eat  10  acres  in  9  weeks,  how  many  oxen  would  it  require  to 
eat  24  acres  in  18  weeks,  the  grass  to  be  growing  uniformly  ? 

Ans.  36  oxen. 

OPERATION    BY    ANALYSIS. 

Each  ox  eats  a  certain  quantity  in  each  week,  which  we  may 
suppose  to  be  100  pounds  ;  and  of  the  whole  quantity  eaten  in 
each  case,  a  part  must  have  already  grown  during  the  time  of 
eating. 

Then,  by  the  first  conditions  of  the  question, 

12  X  4  X  100  =  48001bs.  =  whole  quantity  on  3£  acres  for 
4  weeks. 
'  4800  -T-  3^  =  14401bs.  =  whole  quantity  on  1  acre  for  4  weeks. 

By  the  second  conditions  of  the  question, 

21  X  9  X  1QQ  —  189001bs.  =  whole  quantity  on  10  acres  for 
9  weeks. 

18900  -;-  10  =  18901bs.  =  whole  quantity  on  1  acre  for  9 
weeks. 

1890  —  1440  —  4501bs.  =  the  quantity  grown  on  an  acre  for 
9  —  4  =  5  weeks. 

450  -r-  9  —  4  =  901bs.  =  the  quantity  which  grows  on  each 
acre  for  1  week. 

90  X  3^  X  4  —  12001bs.  =  quantity  grown  on  3£  acres  for  4 
weeks. 

4800  —  1200  =  36001bs.  =  original  quantity  of  grass  on  3£ 
acres. 

3600  -r-  3£  =  lOSOlbs.  —  original  quantity  on  1  acre. 

Then,  by  the  last  condition  of  the  question, 

24  x  1080  =  259201bs.  =  original  quantity  on  24  acres. 

24  x  90  x  18  —  388801bs.  =  quantity  which  grows  on  24 
acres  in  18  weeks. 

25920  -f-  38880  =  648001bs.  =  whole  quantity  on  24  acres 
for  18  weeks. 

64800  -»  18  =  36001bs.  =  quantity  to  be  eat  from  24  acres 
each  week. 

3600  -H  100  =  36  =  number  of  oxen   required  to  eat  the 
whole,  and  the  answer  to  the  question. 
25 


290  EXCHANGE.  [SECT  LXXIII. 

9.  There  is  a  fish  whose  head  weighs*  15  pounds,  his  tail 
weighs  as  much  as  his  head  and  £  as  much  as  his  body,  and  his 
body  weighs  as   much  as  his  head  and  tail.     What  was  the 
weight  of  the  fish  ?  Ans.  721bs. 

10.  Suppose  a  clock  to  have  an  hour-hand,  a  minute-hand, 
and  a  second-hand,  all  turning  on  the  same  centre.     At   12 
o'clock  all  the  hands  are  together  ancf  point  at  12. 

(1.)  How  long  will  it  be  before  the  second-hand  will  be  be- 
tween the  other  two  hands,  and  at  equal  distances  from  each  ? 

Ans.  60-j^V  seconds. 

(2.)  Also  before  the  minute-hand  will  be  equally  distant  be- 
tween the  other  two  hands?  Ans.  61f§^  seconds. 

(3.)  Also  before  the  hour-hand  will  be  equally  distant  be- 
tween the  other  two  hands  ?  Ans.  59        seconds. 


SECTION  LXXIII. 
EXCHANGE. 

EXCHANGE  is  the  act  of  paying  or  receiving  the  money  of  one 
country  for  its  equivalent  in  the  money  of  apother  country,  by 
means  of  Bills  of  Exchange.  This  operation,  therefore,  com- 
prehends both  the  reduction  of  moneys  and  the  negotiation  of 
bills.  It  determines  the  comparative  value  of  the  currencies  of 
all  nations,  and  shows  how  foreign  debts  are  discharged,  loans 
and  subsidies  paid,  and  other  remittances  made  from  one  coun- 
try to  another,  without  the  risk,  trouble,  or  expense  of  transport- 
ing specie  or  bullion. 

BILLS  OF  EXCHANGE. 

A  Bill  of  Exchange  is  a  written  order  for  the  payment  of  a 
certain  sum  of  money,  at  an  appointed  time.  It  is  a  mercantile 
contract,  in  which  four  persons  are  mostly  concerned ;  viz. 

1.  The  drawer,  who  receives  the  value,  and  is  also  called  the 
maker  and  seller  of  the  bill. 

2.  The  person  upon  whom  the  bill  is  drawn  is  called  the 
drawee.     He  is  also  called  the  acceptor,  when  he  accepts  the 
bill,  which  is  an  engagement  to  pay  it  when  due. 

3.  The  person  who  gives  value  for  the  bill,  who  is  called  the 
buyer,  taker,  and  remitter. 

4.  The  person  to  whom  it  is  ordered  to  be  paid,  who  is  called 


SECT.  LXX.III.]  EXCHANGE.  291 

the  payee,  and  who  may,  by  indorsement,  pass  it  to  any  other 
person. 

Most  mercantile  payments  are  made  in  Bills  of  Exchange, 
which  generally  pass  from  hand  to  hand,  until  due,  like  any 
other  circulating  medium  ;  and  the  person  who  at  any  time  has 
a  bill  in  his  possession  is  called  the  holder. 

When  the  holder  of  a  bill  disposes  of  it,  he  writes  his  name 
on  the  back,  which  is  called  indorsing ;  and  the  payee  should 
be  the  first  indorse r.  If  the  bill  be  indorsed  in  favor  of  any 
particular  person,  it  is  called  a  special  indorsement ;  and  the 
person  to  whom  it  is  thus  made  payable  is  called  the  indorsee, 
who  must  also  indorse  the  bill  if  he  negotiates  it.  Any  person 
may  indorse  a  bill,  and  every  indorser  (as  well  as  the  acceptor, 
or  payee)  is  a  security  for  the  bill,  and  may  therefore  be  sued 
for  payment. 

The  term  of  a  bill  varies  according  to  the  agreement  between 
the  parties,  or  the  custom  of  countries.  Some  bills  are  drawn 
at  sight ;  others,  at  a  certain  number  of  days,  or  months,  after 
sight  or  after  date  ;  and  some,  at  usance,  which  is  the  customa- 
ry or  usual  term  between  different  places.  te 

Days  of  grace  are  a  certain  number  of  days  granted  to  the 
acceptor,  after  the  term  of  a  bill  is  expired.  Three  days  are 
usually  allowed. 

In  reckoning  when  a  bill,  payable  after  date,  becomes  due, 
the  day  on  which  it  is  dated  is  not  included  ;  and  if  it  be  a  bill 
payable  after  sight,  the  day  of  presentment  is  not  included. 
When  the  term  is  expressed  in  months,  calendar  months  are 
understood ;  and  when  a  month  is  longer  than  the  preceding,  it 
is  a  rule  not  to  go  in  the  computation  into  a  third  month. 

Thus,  if  a  bill  be  dated  the  28th,  29th,  30th,  or  31st  of  Janu- 
ary, and  payable  one  month  after  date,  the  term  equally  expires 
on  the  last  day  of  February,  to  which  the  days  of  grace  must, 
of  course,  be  added ;  and  therefore  the  bill  becomes  due  on 
the  3d<of  March. 

Form  of  a  Bill  of  Exchange. 

Boston,  September  25,  1835. 
Exchange  for  £  5,000  sterling. 

At  ninety  days'  sight  of  this,  my  first  Bill  of  Exchange  (sec- 
ond and  third  of  the  same  tenor  and  date  unpaid),  pay  to  James 
Aye/,  or  order,  five  thousand  pounds  sterling,  with  or  without 


John  L.  French. 
Messrs.  Dana  &  Hyde, 
Merchants,  Liverpool. 


292  EXCHANGE.  [SECT.  LXXIII. 

ACCEPTING  BILLS. 

When  a  bill  is  presented  for  acceptance,  it  is  generally  left 
till  the  next  day  ;  and  the  common  way  of  accepting  is  for  the 
drawee  to  write  his  name  at  the  bottom,  or  across  the  body  of 
the  bill,  with  the  word  accepted. 

When  two  or  more  persons  are  in  partnership,  the  accept- 
ance of  one  binds  all  the  others,  if  the  bill  concerns  their  joint 
trade  •;  but  if  it  should  be  made  known  to  the  person  who  re- 
ceives the  bill  that  it  concerns  the  acceptor  only,  in  a  distinct 
interest,  he  alone,  as  acceptor,  can  be  sued. 

A  clerk  or  servant  may  accept  a  bill  for  his  master,  when  he 
has  authority  for  that  purpose  ;  or  if  he  usually  transacts  busi- 
ness of  this  nature  for  him  ;  and  his  acceptance  binds  the  mas- 
ter. But  if  the  bill  be  drawn  nominally  on  the  servant,  direct- 
ing him  to  place  it  to  the  account  of  his  master,  and  if  the  ser- 
vant should  accept  it  generally,  without  specifying  that  he  does 
it  for  his  master's  account,  the  acceptance  binds  the  servant 
only,  and  not  his  employer. 

When  a  bill  is  drawn  for  the  account  of  a  third  person,  and 
is  accepted  as  such,  and  he  fails  without  making  provision  for 
its  payment,  the  acceptor  must  discharge  the  bill,  and  can  have 
no  recourse  against  the  drawer. 

A  bill  may  be  accepted  to  be  paid  at  a  longer  period  than  is 
mentioned  in  the  bill,  or  to  pay  a  part  of  the  sum  only ;  such 
an  acceptance  is  binding  on  him  who  made  it ;  but  the  holder  is 
at  liberty  to  take  it  as  it  is  offered,  or  to  act  as  if  acceptance 
had  been  entirely  refused. 

INDORSING  BILLS. 

Bills  payable  to  bearer  are  transferred  by  simple  delivery, 
and  without  any  indorsement ;  but  in  order  to  transfer  a  bill 
payable  to  order,  the  holder  must  express  his  order  of  paying 
to  another  person,  which  is  always  done  by  an  indorsement. 

An  indorsement  may  be  blank  or  special.  A  Hank  indorse- 
ment consists  only  of  the  indorsees  name,  and  the  bill  becomes 
then  transferable  by  simple  delivery;  a  special  indorsement 
orders  the  money  to  be  paid  to  some  particular  person,  or  to 
his  order ;  a  blank  indorsement  may  always  be  filled  up  with 
any  person's  name,  so  as  to  make  it  special. 

An  indorsement  may  take  place  at  any  time  after  the  bill  is 
issued,  even  after  the  day  of  payment  is  elapsed. 


SECT.  LX.XIII.]  EXCHANGE.  293 

A  person  who  receives  a  bill  with  a  blank  indorsement  may 
take  it  as  indorsee,  negotiate  it  again,  or  demand  payment  on  his 
own  account,  or  he  may  receive  the  money  as  agent,  or  for  the 
account  of  the  indorser ;  and  the  latter,  notwithstanding  his  in- 
dorsement, may  still  appear  as  holder  in  an  action  against  the 
drawer  or  acceptor. 

A  special  indorsement  need  not  contain  the  words  to  order, 
and  the  bill  is  negotiable  ;  it  may  also  be  restrictive,  giving  au- 
thority to  the  indorsee  to  receive  the  money  for  the  indorser, 
but  not  to  transfer  the  bill  again  to  another. 

An  indorsement  for  part  of  the  money  only  is  not  valid,  ex- 
cept with  regard  to  him  who  makes  it.  The  drawer  and  ac- 
ceptor are  not  bound  by  it. 

When  the  holder  of  a  bill  dies,  his  executors  may  indorse  it ; 
but  by  so  doing  they  become  answerable  to  their  indorsee 
personally,  and  not  as  executors. 

PROTESTING  BILLS. 

When  acceptance  or  payment  has  been  refused,  the  holder 
of  the  bill  should  give  regular  and  immediate  notice  to  all  the 
parties  to  whom  he  intends  to  resort  for  payment ;  and  if,  on 
account  of  unnecessary  delay,  a  loss  should  be  incurred  by  the 
failure  of  any  of  the  parties,  the  holder  must  bear  the  loss. 

With  respect  to  the  manner  in  which  notices  of  non-accept- 
ance or  non-payment  are  to  be  given,  a  difference  exists  be- 
tween inland  and  foreign  bills. 

For  foreign  bills  a  protest  is  indispensably  necessary ;  thus, 
a  public  notary  is  to  appear  with  the  bill,  and  to  demand  either 
acceptance  or  payment ;  and,  on  being  refused,  he  is  to  draw 
up  an  instrument,  called  a  protest,  expressing  that  acceptance 
or  payment  has  been  demanded  and  refused,  and  that  the  hold- 
er of  the  bill  intends  to  recover  any  damages  which  he  may 
sustain  in  consequence.  This  instrument  is  admitted  in  foreign 
countries  as  legal  .proof  of  the  fact. 

It  is  customary,  as  a  precaution  against  accidents  or  miscar- 
riage, to  draw  three  copies  of  a  foreign  bill,  and  to  send  them 
by  different  posts.  They  are  denominated  the  first,  second, 
and.  third  of  exchange ;  and  when  any  one  of  them  is  paid, 
the  rest  become  void  and  of  no  value.  When  the  acceptor  of 
a  bill  becomes  insolvent,  or  absconds  before  the  term  of  pay- 
ment is  expired,  the  holder  may  cause  a  notary  to  demand  bet- 
ter security,  and,  on  that  being  refused,  to  protest  the  bill  for 
25* 


294  EXCHANGE.  [SECT.  LXXIII. 

want  of  it.     In  such  cases,  however,  the  most  general  practice 
is  to  wait  the  regular  time,  till  the  bill  becomes  due. 

The  damages  incurred  by  non-acceptance  and  non-payment, 
besides  interest,  consist  usually  of  the  exchange,  or  reexchange, 
commission,  and  postage  together,  with  the  expenses  of  protest, 
and  interest.  The  exchange  is  reckoned  according  to  the 
course  at  sight  from  the  place  where  the  protest  is  made  to 
the  place  where  the  bill  is  to  be  paid  by  the  drawer;  and,  if  it 
be  not  paid  there,  the  exchange  is  then  reckoned  from  the  same 
place  to  that  where  the  bill  is  paid,  and  also  double  commis- 
sion. The  interest  commences  from  the  day  when  the  demand 
was  made. 

RECOVERING   BILLS. 

The  drawer,  acceptor,  and  even  indorser  of  a  bill  are  equal- 
ly liable  to  the  payment  of  it ;  and  though  the  holder  can  have 
but  one  satisfaction,  yet,  until  such  satisfaction  is  actually  had, 
he  may  sue  any  of  them,  or  all  of  them,  either  at  the  same 
time  or  in  succession,  and  obtain  judgment  against  them  all, 
till  satisfaction  be  made.  Proceedings  cannot  be  staid  in  any 
action  except  on  payment  of  the  debt  and  costs,  not  only  in  that 
action,  but  in  all  the  others  in  which  judgment  has  not  been 
obtained  ;  and  though  the  principal  sum  should  be  paid  by  one 
of  the  parties,  still  costs  may  be  recovered  in  the  several  ac- 
tions against  the  others. 

When  acceptance  is  refused,  and  the  bill  is  returned  by  pro- 
test, an  action  may  be  commenced  immediately  against  the 
drawer,  though  the  regular  time  of  payment  be  not  arrived. 
His  debt,  in  such  a  case,  is  considered  as  contracted  the  mo- 
ment the  bill  is  drawn.  Thus,  if  before  the  bill  is  returned  the 
drawer  should  become  a  bankrupt,  the  debt  was  contracted  be- 
fore the  commission  of  bankruptcy  took  place. 

Nothing  will  discharge  an  indorser  from  his  engagement  but 
the  absolute  payment  of  the  money ;  not  even  a  judgment  re- 
covered against  the  drawer,  or  any  previous  indorser,  or  any 
execution  against  any  of  them,  unless  the  money  be  paid  in 
consequence. 

INLAND  EXCHANGE,  OR  DRAFTS. 

By  Inland  Exchange  is  understood  the  act  of  remitting  bills 
to  places  in  the  same  country  ;  by  which  means  debts  are  dis- 
charged more  conveniently  than  by  cash  remittances. 


SECT.  LXXIII.]  EXCHANGE.  295 

Suppose,  for  example,  A,  of  Boston,  is  creditor  to  B,  of  Balti- 
more, $100,  and  C,  of  Boston,  debtor  to  D,  of  Baltimore,  $  100, 
both  these  debts  may  be  discharged  by  means  of  one  bill. 
Thus,  A  draws  for  this  sum  on  B,  and  sells  his  bill  to  C,  who 
remits  it  10  D,  and  the  latter  receives  the  amount,  when  due, 
from  B.  Here,  by  a  transfer  of  claims,  the  Boston  debtor  pays 
the  Boston  creditor;  and  the  Baltimore  debtor  the  Baltimore 
creditor ;  and  no  money  is  sent  from  one  place  to  the  other. 
The  same  would  take  place  if  D,  of  Baltimore,  drew  on  C,  of 
Boston,  and  sold  his  bill  to  B,  of  Baltimore,  who  should  send  it 
to  A,  of  Boston ;  the  effect,  in  either  case,  being  merely  a 
transfer  of  debtors  and  creditors. 

NOTE. —  In  this  operation,  A  is  the  drawer  and  seller-, ,  B  the  drawee 
and  acceptor,  C  the  buyer  and  remitter,  and  D  the  payee,  if  his  name  be 
mentioned  in  the  bill  ;  and  he  is  the  holder  when  he  receives  the  bill  from 
A.  When  D,  or  any  other  holder,  presents  the  bill  for  acceptance  or 
payment,  he  is  called  the  presenter. 

By  the  foregoing  example,  it  appears  that  reciprocal  and 
equal  debts  due  between  two  places  may  be  discharged  without 
remitting  specie  ;  and  it  may  be  supposed  that  such  an  opera- 
tion is  of  equal  convenience  to  all  parties  concerned  ;  but  when 
the  debts  are  unequal,  the  advantage  must  be  different,  as  the 
obligation  of  remittance  is  no  longer  mutual,  because  the  debt- 
or place  must  pay  its  balance,  either  by  sending  cash  or  bills; 
and  as  the  latter  mode  is  generally  preferred,  an  increased  de- 
mand for  bills  must  be  the  consequence,  which  enhances  their 
price,  as  it  would  that  of  any  other  article  of  sale  or  purchase. 

PAR  OF  EXCHANGE. 

The  Par  of  Exchange  may  be  considered  under  two  gen- 
eral heads  ;  viz.  the  intrinsic  par  and  the  commercial  par,  each 
of  which  admits  of  subordinate  divisions  and  distinctions. 

The  intrinsic  par  is  the  value  of  the  money  of  one  country, 
compared  with  that  of  another,  with  respect  both  to  weight  and 
fineness. 

The  commercial  par  is  the  comparative  value  of  the  moneys 
of  different  countries,  according  to  the  weight,  fineness,  and 
market  prices  of  the  metals. 

Thus,  two  sums  of  different  countries  are  intrinsically  at 
par,  when  they  contain  an  equal  quantity  of  the  same  kind  of 
pure  metal ;  and  two  sums  of  different  countries  are  commer- 
cially at  par,  when  they  can  purchase  an  equal  quantity  of  the 
same  kind  of  pure  metal. 


296  EXCHANGE.  [SECT.  LXXIII. 

COURSE   OF   EXCHANGE. 

The  Course  of  Exchange  is  the  variable  price  of  the  money 
of  one  country,  which  is  given  for  a  fixed  sum  of  money  of  an- 
other country  ;  the  latter  is  called  the  certain,  and  the  former 
the  uncertain  price,  as  before  stated. 

When  the  market  price  of  foreign  bills  is  above  par,  the  ex- 
change is  said  to  be  favorable  to  the  place  that  gives  the  cer- 
tain for  the  uncertain. 

It  should,  however,  be  recollected,  that  when  the  exchange  is 
favorable  to  a  place,  it  is  only  so  to  the  buyers  and  remitters  of 
bills,  but  it  is  unfavorable  to  the  drawers  and  sellers. 

Thus,  the  interest  of  the  remitter  is  identified  with  that  of  the 
place  where  he  purchases  the  bill,  and  the  interest  of  the  draw- 
er with  that  of  the  place  where  his  funds  are  established,  and 
on  which  he  draws. 

It  is  natural  to  inquire  why  such  prices  are  considered  favor- 
able or  unfavorable,  if  the  drawers  and  remitters,  whose  inter- 
ests are  opposite,  are  natives  of  the  same  country.  The  usual 
answer  is,  that  when  the  exchange  is  against  a  place,  it  becomes 
the  interest  of  remitters  to  pay  their  foreign  debts  in  specie  in- 
stead of  bills,  and  the  exportation  of  the  precious  metals  is  often 
considered  a  national  disadvantage. 

The  fluctuations  of  exchange  are  occasioned  by  various  cir- 
cumstances, both  political  and  commercial.  The  principal 
cause  is  generally  stated  to  be  the  balance  of  trade  ;  that  is,  the 
difference  between  the  commercial  exports  and  imports  of  any 
one  country  with  respect  to  another.  Experience,  however, 
shows,  that  the  exchange  may  be  unfavorable  to  a  country, 
when  the  balance  of  trade  is  greatly  in  its  favor ;  for  the  de- 
mand for  bills  must  chiefly  depend  on  the  balance  of  such  debts 
as  come  into  immediate  liquidation  ;  that  is  to  say,  on  the  bal- 
ance of  payments. 

Besides,  it  does  not  follow  that  large  exports  are  always  suc- 
cessful, or  quick  in  their  returns ;  and  even  should  it  be  the 
case,  the  balance  of  payments  may  still  be  unfavorable  from 
political  causes. 

When  any  alteration  takes  place  in  the  coin  or  currency  of 
a  country,  the  exchange  will,  of  course,  vary  so  as  to  keep 
pace  or  correspond  to  such  alteration.  This,  however,  cannot 
be  considered  a  change  in  the  price  of  bills,  but  in  the  money 
in  which  they  are  bought  or  sold. 

In  times  of  peace  the  course  of  exchange  seldom  remains 


SECT.  LXXIII.]  EXCHANGE.  297 

long  unfavorable  to  any  country,  at  least,  beyond  the  expense 
that  might  be  incurred  by  the  transportation  of  the  precious 
metals  ;  for  bullion  is  considered  the  universal  currency  of 
merchants,  and  exchange  gives  it  circulation,  and  thus  tends  to 
maintain  the  level  of  money  throughout  the  commercial  world. 

An  unfavorable  course  of  exchange  may,  therefore,  be  cor- 
rected, either  by  the  exportation  of  bullion,  or  the  shipment  of 
goods,  —  and  another  method  sometimes  offers,  by  negotiating 
bills  through  several  places  ;  but  the  latter  remedy  must  fail  if 
the  exchange  be  universally  unfavorable. 

From  what  has  been  said  of  the  causes,  both  commercial 
and  political,  which  produce  the  fluctuations  of  exchange,  and 
which  sometimes  counteract  or  balance  each  other,  the  follow- 
ing simple  conclusion  may  be  drawn;  —  that  bills  rise  or  fall 
in  their  prices,  like  any  other  salable  articles,  according  to  the 
proportion  that  exists  between  the  demand  and  supply. 


GREAT   BRITAIN 

Accounts  are  kept  in  Great  Britain  in  pounds,  shillings,  pence, 
and  farthings,  sterling.  The  principal  coins  are  guineas,  sov- 
ereigns, crowns,  and  their  fractional  parts. 

The  Guinea      =21  Shillings,  sterling,  =  $  5.07,5 
"     Sovereign  =  20          "  "       =    4.84,6 

"     Crown       =5          "  "       =     1.08 

The  pound  sterling,  or  sovereign,  has  different  values  accord- 
ing to  circumstances. 

The  exchange  value  is  $  4.44|. 

Its  legal  value  at  the  mint  is  $  4.86,6. 

It  is  received  at  the  custom-house  in  payment  of  duties  at 


Its  commercial  value  is  from  $  4.82  to  $  4.86  ;  and  its  value 
is  often  greater  in  some  of  the  States  than  in  others.  The  price 
of  foreign  coin  in  our  market  determines  its  value. 

The  commercial  value  is  generally  about  9  per  cent,  more 
than  the  exchange  value. 

Thus,  the  exchange  value  being  =  $  4.44f- 

To  which  we  add  9  per  cent,  premium  =        .40 
The  commercial  value  will  be  =  $  4.84f 

EXAMPLES. 
1.  What  must  a  merchant  in  Boston  pay  in  dollars  and  cents 


298  EXCHANGE.  [SECT.  LXXIII. 

for  a  bill  of  9765<£.  15s.  6d.  on  Liverpool,  the  premium  being 
9  per  cent.  ?  Ans.  $  47,309.75+. 

2.  Kimball,  Jewett,  &  Co.,  of  Boston,  wish  to  purchase  a  bill 
of  1876 !<£.  10s.  on  Liverpool,  the  premium  being  8£  percent. ; 
what  will  be  the  cost  of  the  bill  ?     Ans.  20,356£.  4s.  6£d.+. 

3.  If  I  pay  94-  per  cent,  premium,  what  will  be  the  amount 
of  a  bill  on  London  which  I  can  purchase  for  $  81,727.75? 

Ans, 

4.  John  Jones,  of  Boston,  has  consigned  a  cargo  of  flour, 
valued  at  17,000^.,  to  Robert  Morrison,  Liverpool.    R.  S.  Davis, 
of  Boston,  who  is  about  to  import  many  valuable  books,  has 
purchased  of  J.  Jones  a  bill  of  exchange,  at  6  per  cent,  premi- 
um, for  the  value  of  the  above  flour.     The  following  is  the 
form  of  the  bill :  — 

Exchange  for  17,000 «£.  Boston,  August  2d,  1847. 

Ninety  days  after  sight  of  this  my  first  Bill  of  Exchange 
(second  and  third  of  the  same  tenor  and  date  unpaid),  pay  to 
Robert  S.  Davis,  or  order,  seventeen  thousand  pounds  sterling, 
with  or  without  further  advice.  J  h  J 

Robert  Morrison,  Esq., 
Merchant,  Liverpool. 

What  should  be  paid  for  the  above  bill  ? 

Ans.  880,088.88+. 

FRANCE. 

Accounts  in  France  are  kept  in  francs  and  centimes.  The 
mercantile  value  of  some  of  the  principal  coins  of  France,  in 
United  States  currency,  is  as  follows  :  — 

The  Double  Napoleon  or  Louis,  40  francs,  =  $  7.44 
"     Napoleon  or  Louis,  20  francs,  =     3.72 

"     Franc,  100  centimes,  =       .18f 

5.  If  a  merchant  in  Boston  should  remit  to  Paris  172,000 
francs,  exchange  being  1£  per  cent.,  what  will  be  the  cost  of 
his  bill  in  United  States  currency  ?  Ans.  $32,471.88. 

6.  What  must  a  merchant  in  New  York  pay  for  a  bill  on 
Havre  for  76,000  francs,  the  exchange  being  5  francs  8  cen- 
times for  a  dollar  ?  Ans.  $  14,960.62+. 

7.  What  is  the  value  of  a  bill  on  Paris  for  79,000  francs, 
exchange  being  2  per  cent,  below  par  ?'       Ans.  $14,400.12. 

8.  John  Smith,  of  Boston,  remits  to  Bordeaux  $17,280,  the 
exchange  being  5  francs  10  centimes  for  a  dollar.     Required 
the  amount  in  francs.  Ans.  88,128  francs. 


SECT.  I.XXTII.]  EXCHANGE.  299 

AMSTERDAM. 

Accounts  are  kept  in  florins,  stivers,  and  pfennings,  or  in 
pounds,  shillings,  and  pence  Flemish. 

16  Pfennings  =  1  Stiver. 

20  Stivers  =  1  Florin,  or  Guilder. 

Flemish- 


20  Shillings  Flemish,  or  6  Florins,    =  1  Pound  Flemish. 
2£  Florins,  or  50  Stivers,  =  1  Rix  Dollar. 

9.  United  States  on  Amsterdam.      Reduce  896   florins   10 
stivers  to  U.  S.  money,  exchange  at  38  cents  per  florin. 

Ans.  $  340.67. 

10.  Amsterdam  on  the  United  States.     Reduce  8  340.67  to 
the  money  of  Amsterdam,  exchange  at  38  cents  per  florin. 

Ans.  896  florins  10  stivers. 

CONSTANTINOPLE. 

Accounts  are  kept  in  piasters,  paras,  and  aspers,  or  in  pias- 
ters and  aspers  ;  sometimes  in  piasters  and  half-paras,  or  in 
piasters  and  minas. 

3  Aspers  =  1  Para. 

40  Paras  =  1  Piaster,  or  Turkish  Dollar. 

Also,  80  Half-paras  or  100  Minas  =  1  Piaster. 

11.  United  States  on  Constantinople.     Reduce  78  piasters 
20  paras  to  U.  S.  money,  exchange  at  40  cents  per  piaster. 

Ans.  $31.40. 

12.  Constantinople  on  tfle  United  States.      Reduce  $31.40 
to  the  money  of  Constantinople.      Ans.  78  piasters  20  paras. 

COPENHAGEN. 

Accounts  are  kept  here  in  rix  dollars,  marks,  and  skillings 
Danish,  but  sometimes  in  rix  dollars,  marks,  and  skillings  lubs. 
Pfennings  are  also  occasionally  reckoned. 
12  Pfennings  =  1  Skilling. 

16  Skillings  =  1  Mark. 

$  Marks  Danish,  or  3  Marks  Lubs  =  1  Ryksdaler,  or  Rix  Dollar. 

13.  United  States  on  Copenhagen.     Reduce  896  rix  dollars 
3  marks  to  U.  S.  money,  exchange  at  50  cents  per  rix  dollar. 

Ans.  $  448.25. 


300  EXCHANGE.  [SECT.  LXXIII. 


DANTZIC. 

Accounts  are  computed  here  in  florins,  groschen,  and  pfen- 
nings. 

3  Pfennings  =  1  Groschen. 
30  Groschen  =  1  Florin. 
3  Florins       =  1  Rix  Dollar. 

14.  United  States  on  Dantzic.      Reduce  196  rix  dollars  2 
florins  to  U.  S.  money,  exchange  at  17  cents  per  florin. 

Ans.  9  100.30. 

15.  Dantzic  on  the  United  States.     Reduce  $  100.30  to  the 
currency  of  Dantzic.  Ans.  196  rix  dollars  2  florins. 


HAMBURG. 

Computations  are  made  in  marks,  schillings,  and  pfennings, 
banco  or  current ;  banco  bears  an  agio  on  currency  of  from  20 
to  25  per  cent. 

12  Pfennings  =  1  Schilling,  or  Sol,  Lubs. 

16  Schillings  Lubs  =  1  Mark. 
3  Marks  =  1  Rix  Dollar. 

16.  United  States  on  Hamburg.     Reduce  675  rix  dollars  2 
marks  to  U.  S.  money,  exchange  at  30  cents  per  mark. 

Ans.  $608.10. 

17.  Hamburg  on  the  United  States.     Reduce  $608.10   to 
the  currency  of  Hamburg,  exchange  at  30  cents  per  mark. 

Ans.  675  rix  dollars,  2  marks. 


LEGHORN. 

Accounts  are  kept  in  pezze,  soldi,  and  denari  di  pezza. 

12  Denari  di  Pezza  =  1  Soldo  di  Pezza. 
20  Soldi  di  Pezza     =  1  Pezza. 
12  Denari  di  Lira    =  1  Soldo  di  Lira. 
20  Soldi  di  Lira       =  1  Lira. 

18.  United  States  on  Leghorn.     Reduce  286  pezze  10  soldi 
to  U.  S.  money,  exchange  at  90  cents  per  pezza. 

Ans.  S  257.85. 

19.  Leghorn  on  the  United  States.     Change  $  257.85  to  the 
currency  of  Leghorn.  Ans.  286  pezze  10  soldi." 


SECT.  LXXIII.]  EXCHANGE.  301 

MILAN. 

Accounts  are  kept  in  lire,  soldi,  and  denari  correnti  or  impe- 
riali. 

12  Denari  =      1  Soldo. 

20  Soldi  =      1  Lira.     • 

106  Soldi  or  Lire  Imperial!  =150  Soldi  or  Lire  Correnti. 
150  Soldi  or  Lire  Correnti  =      1  Filippo. 
117  Soldi  Imperiali  =      1  Scudo  or  Crown. 

20.  United  States  on  Milan.     Change  176  lire  10  soldi  to 
U.  S.  money,  the  lira  being  valued  at  20  cents.     Ans.  $  35.30. 

21.  Milan  on  the  United  States.     Reduce  $  35.30  to  the  cur- 
rency of  Milan.  Ans.  176  lire  10  soldi. 

NAPLES. 

Computations  are  made  in  ducats  of  100  grains. 
10  Grani   =  1  Carlino. 
10  Carlini  =  1  Ducato  di  Regno. 

22.  United  States  on  Naples.     Change  769  ducati  di  regno 
5  carlini  to  United  States  money,  the  value  of  the  ducato  being 
80  cents.  Ans.  $615.60. 

23.  Naples  on  the  United  States.     Reduce  $615.60  to  the 
currency  of  Naples.  Ans.  769  ducati  di  regno  5  carlini. 

SICILY. 

Accounts  are  kept  here  in  oncie,  tari,  and  grani ;  and  also  in 
scudi,  tari,  and  grani. 

20  Grani  =  1  Taro. 
30  Tari    =  1  Oncia. 

Also,  12  Tari    =  1  Scudo  or  Sicilian  Crown. 
5  Scudi  =  2  Oncie. 

24.  United  States  on  Sicily.     Change  876  oncie  3  scudi  to 
U.  S.  money,  the  oncia  being  valued  at  $  2.38,3. 

Ans.  8  2090.36,7f . 

25.  Sicily  on  the  United  States.   Change  $  2090.36,7f  to  the 
currency  of  Sicily.  Ans.  876  oncie  3  scudi. 

RUSSIA. 

Computations  are  made  here  in  rubles  and  kopecks. 
10  Kopecks  =  1  Grieve  or  Grievener. 
10  Grieves  =  1  Ruble. 
26 


302  EXCHANGE.  [SECT.  LXXHI. 

26.  United  States  on  Russia.     What  is  the  value  of  7684  ru- 
bles 8  grieves  in  U.  S.  money,  the  value  of  the  ruble  in  the 
United  States  being  75  cents  ?  Ans.  $  5763.60. 

27.  Russia  on  the   United  States.     What  is  the  value  of 
$  5763.60  in  Russian  currency?      Ans.  7684  rubles  8  grieves. 

ROME.  ' 

Accounts  are  kept  in  scudi  moneta  and  baiocchi ;  or  in  scudi 
di  stampa  soldi  and  denari  d'  oro ;  quattrini  and  mezzi  quattrini 
are  sometimes  reckoned. 

2  Mezzi  Quattrini  =  1  Quattrino. 
5  Quattrini  =  1  Baioccho. 

10  Baiocchi  =  1  Paolo. 

10  Paoli  =  1  Scudo  Moneta,  or  Roman  Crown. 

28.  United  States  on  Rome.     Change  7689  scudi  moneta  to 
U.  S.  money,  the  value  of  the  scudo  being  8  1.00,0^. 

Ans.  $7694. 15TWa. 

29.  Rome  on  the  United  States.     Change  $7694.15Ty$r  to 
Roman  currency.  Ans.  7689  scudi  moneta. 

SPAIN. 

The  general  mode  of  keeping  accounts  in  Spain  is  in  mara- 
vedis  and  reals. 

34  Maravedis  =  1  Real. 

8  Reals  =  1  Dollar  of  Plate. 

375  Maravedis  =  1  Ducat  of  Exchange. 

4  Dollars  of  Plate  =  1  Pistole  of  Exchange. 

Exchanges  are  generally  computed  in  denominations  of  plate, 
which  is  always  understood  to  be  old  plate,  if  new  plate  be  not 
mentioned. 

There  are  three  principal  denominations  of  these  imaginary 
moneys  in  which  exchanges  are  generally  transacted  ;  viz., 
dollars,  doubloons,  and  ducats,  and  they  are  divided  into  reals 
and  maravedis  of  plate,  and  sometimes  converted  into  vellon 
and  other  denominations. 

The  dollar  of  exchange,  also  called  peso  or  piastre  de  cambio, 
or  de  plata,  is  divided  into  8  reals  of  34  maravedis  of  plate 
each,  and  sometimes  into  16  quartos. 

The  doubloon  de  plata,  or  pistole  of  exchange,  is  four  times 
the  dollar,  and  therefore  contains  32  reals,  or  1088  maravedis 
of  plate. 


SECT.  LXXIII.]  EXCHANGE.  303 

The  ducat  of  plate,  also  called  ducado  de  cambio,  contains  11 
reals  1  maravedi,  or  375  maravedis  of  plate. 

At  Alicant,  Valencia,  and  Barcelona,  exchanges  are  transact- 
ed in  libras  of  20  sueldos,  or  240  dineros. 

The  libra  of  Alicant  and  Valencia  is  the  dollar  of  plate.  This 
is  sometimes  divided  into  10  reals  of  new  plate,  which  are,  of 
course;  equal  to  8  reals  of  old  plate. 

The  libra  of  Barcelona,  commonly  called  libra  Catalan,  is 
worth  5|-  reals  of  plate ;  hence  7  of  those  libras  equal  5  dol- 
lars of  plate,  and  therefore  28  sueldos  Catalan  equal  1  dollar. 

The  hard  dollar  of  20  reals  vellon  is  occasionally  used  in 
exchanges,  and  is  also  divided  into  12  reals,  each  of  16  quartos. 
The  current  dollar,  which  is  an  imaginary  money,  valued  at 
two  thirds  of  the  hard  dollar,  is  divided  into  8  reals,  and  the 
real  into  16  quartos.  The  two  latter  are  the  principal  moneys 
of  exchange  used  at  Gibraltar. 

30.  United  States  on  Spain.     What  is  the  value  in  United 
States  money  of  7600  dollars  of  plate,  exchange  at  75  cents  for 
a  plate  dollar  ?  Ans.  $  5700. 

31.  Spain  on  the  United  States.  What  is  the  value  of  $  5700 
in  Spanish  money,  exchange  at  75  cents  per  dollar  of  plate  ? 

Ans.  7600  dollars  of  plate. 

SWEDEN. 

Accounts  are  kept  in  rix  dollars,  skillings,  and  pfennings. 
12  Pfennings  =  1  Skilling. 
48  Skillings    =  1  Rix  Dollar. 

32.  United  States  on  Sweden.     Reduce  476  rix  dollars  24 
skillings  to  U.  S.  money,  the  rix  dollar  being  valued  at  107 
cents.  Ans.  $509.85,5. 

33.  Sweden  on  the  United  States.    Change  $  509.85,5  to  the 
currency  of  Sweden.  Ans.  476  rix  dollars  24  skillings. 

TURIN. 

Accounts  are  kept  in  lire,  soldi,  and  denari. 
12  Denari  =  1  Soldo. 
20  Soldi     =  1  Lira. 

34.  United  States  on  Turin.     Change  462  lire    10  soldi  to 
U.  S.  money,  exchange  at  20  cents  per  lira.       Ans.  $  92.50. 

35.  Turin  on  the  United  States.    Change  $  92.50  to  the  cur- 
rency of  Turin,  exchange  at  20  cents  per  lira. 

Ans.  462  lire  10  soldi. 


304  EXCHANGE.  [SECT.  LXXIII. 

VIENNA. 

Computations  are   made  in  florins  and  creutzers,  or  in  rix 
dollars  and  creutzers. 

4  Pfennings  =  1  Creutzer. 
60  Creutzers  =  1  Florin. 
1£  Florins     =  1  Rix  Dollar  of  Account. 
2  Florins      =  1      "  "      Specie. 

36.  United  States  on  Vienna.     Reduce  876  rix  dollars,  spe- 
cie, 1  florin  to  U.  S.  money,  the  specie  rix  dollar  being  equal 
to  97  cents.  Ans.  $  850.20,5. 

37.  Vienna  on  the  United  States.     Change  $  850.20,5  to  the 
currency  of  Vienna.      Ans.  876  rix  dollars,  specie,  1  florin. 

EAST  INDIES,  BENGAL,  CALCUTTA,  &c. 

12  Pice  =  1  Anna. 

16  Annas  =  1  Rupee. 

1  Sicca  Rupee  =  2s.  6d.  sterling. 

38.  United  States  on  Calcutta.     Reduce  432  rupees  12  an- 
nas to  U.  S.  money,  exchange  at  50  cents  per  rupee. 

Ans.  $  216.37£. 

39.  Calcutta  on  the  United  States.    Reduce  $  216.37£  to  the 
currency  of  Calcutta,  exchange  at  50  cents  per  rupee. 

Ans.  432  rupees  12  annas. 

BOMBAY. 

100  Rees        =  1  Quarter. 
4  Quarters  =  1  Rupee. 
1  Rupee     =  2s.  4d.  sterling. 

40.  United  States  on  Bombay.     Change  678  rupees  2  quar- 
ters to  U.  S.  money,  the  rupee  being  50  cents. 

Ans.  $  339.25. 

41.  Bombay  on  the  United  States.      Change   $339.25  to 
Bombay  money,  reckoning  the  rupee  at  50  cents. 

Ans.  678  rupees  2  quarters. 

MADRAS. 

Accounts  are  kept  here  in  pagodas,  fanams,  and  cash. 
80  Cash  =  1  Fanam. 

45  Fanams          =  1  Star  Pagoda. 
1  Star  Pagoda  =  8s.  sterling. 


SECT.  LXXIV.J 


VALUE  OF  GOLD  COINS. 


305 


42.  United  States  on  Madras.     Change  375  star  pagodas  to 
U.  S.  money,  the  star  pagoda  being  valued  at  $  l-77£. 

Ans.  $  666.66+. 

43.  Madras  on  the  United  States.     Reduce  $  896  to  the  cur- 
rency of  Madras.  Ans.  504  pagodas. 

TRIESTE. 

Accounts  are  here  kept  in  pfennings,  florins,  and  creutzers. 

4  Pfennings  =  1  Creutzer. 
60  €reutzers  =  1  Florin. 
1£  Florins      =  1  Rix  Dollar. 

44.  United  States  on  Trieste.     What  is  the  value  in  U.  S. 
money  of  769  rix  dollars  40  creutzers,  the  value  of  the   rix 
dollar  being  92  cents  ?  Ans.  $707.88|. 

45.  Trieste  on  the  United  States.     Reduce  $  707.88f  to  the 
currency  of  Trieste.  Ans.  769  rix  dollars  40  creutzers. 


SECTION  LXXIV. 
VALUE   OF   GOLD   COINS, 

ACCORDING  TO  THE  LAWS  OF  MAY  AND  JUNE,  1834 


Names  of  Coins. 

Weight. 

Former 
Standard. 

Standard  of 
July  31st, 
1834. 

UNITED  STATES. 

dwt.    grs. 

$  cts.  m 

$  cts.  m. 

Eagle  coined  before  July  31,  1834, 

11     6 

10.00,0 

10.66,5 

Shares  in  proportion. 

Foreign  Gold. 

AUSTRIAN  DOMINIONS. 

Souverein,          - 

3  14 

3.17,6 

3.37,7 

4  12 

4:29,9 

4.58,9 

Hungarian  Ducat,       - 

2     5| 

2.15,4 

2.29^6 

BAVARIA. 

Carolin,          -_•-.. 

6     5| 

4.64,6 

4.95,7 

Max  d'Or,  or  Maximilian, 

4     4 

3.11,1 

3.31,8 

Ducat,  ------ 

2     5| 

2.13,3 

2.27,5 

BERNE. 

Ducat,  double  in  proportion, 

1  23 

1.85,4 

1.98,6 

Pistole,           

4  21 

4.26,2 

4.54,2 

26* 

306 


VALUE  OF  GOLD   COINS. 


[SECT.  LXXIV. 


BRAZIL. 

dwt.  grs. 

$  cts.  m. 

S  cts.  m. 

Johannes,  half  in  proportion, 

18  00 

16.00,0 

17.06,4 

T^    K  o/-k 

34  12 

30.66,6 

32.70,6 

18     6 

16  22  2 

nQA    1 

Moidore,  half  in  proportion,     - 

6  22 

J.U.«><fe'}« 

6.14,9 

.  *>V7,  1 

6.55,7 

164 

.59,8 

.63,5 

BRUNSWICK. 

Pistole,  double  in  proportion,  - 

4  214 

4.27,1 

4.54,8 

Ducat,       

2     5| 

2.09,2 

2.23,0 

COLOGNE. 

Ducat,  ------ 

2     5| 

2.12,5 

2.26,7 

COLOMBIA. 

Doubloons,         - 

17     9 

14.56,0 

15.53,5 

DENMARK. 

• 

Ducat,  current,       -        -        -        - 

2     0 

1.70,5 

1.81,2 

Ducat,  specie,    - 

2     5| 

2.12,5 

2.26,7 

Christian  d'Or,       -        ... 

4     7 

3.77,0 

4.02,1 

EAST  INDIES. 

Rupee,  Bombay,  1818, 

7  11 

6.65,4 

7.09,6 

Rupee,  Madras,  1818,     -        -        - 

7  12 

6.66,7 

7.11,0 

Pagoda,  Star,     -        -        -        - 

2     4| 

1.68,9 

1.79,8 

ENGLAND. 

Guinea,  half  in  proportion, 

5     8* 

4.79,9 

5.07,5 

Sovereign,          - 

5     24 

4.57,0 

4.84,6 

Seven  Shilling  Piece,     -        -        - 

1  19 

1.60,0 

1.69,8 

FRANCE. 

Double  Louis,  coined  before  1786, 

10  11 

9.08,7 

"9.69,7 

Louis,  coined  before  1786, 

5     54 

4.54,1 

4.84,6 

Double  Louis,  coined  since  1786, 

9  20 

8.59,0 

9.15,3 

Louis,  coined  since  1786, 

4  22 

4.29,5 

4.57,6 

Double  Napoleon,  or  40  francs,   - 
Napoleon,  or  20  francs, 

8     7 
4     34 

7.23,2 
3.61,6 

7.70,2 
3.85,1 

FRANKFORT  ON  THE  MAINE. 

Ducat,  

2     5| 

2.13,7 

2.27,9 

GENEVA. 

Pistole,  old,        ...        - 

4     74 

3.73,7 

3.98,5 

Pistole,  new,          -        -        -        - 

3  155 

3.23,2 

3.44,4 

HAMBURG. 

Ducat,  double  in  proportion, 

2     5| 

2.13,7 

2.27,9 

GENOA. 

0        K3 

2  15  8 

Q  Qn  Q 

HANOVER. 

•      J5 

->  .  L<J  jO 

<6.OVj* 

Double  Geo.  d'Or,  single  in  proportion, 

8  13 

7.48,2 

7.87,9 

SECT.  LXXIV.] 


VALUE  OF  GOLD  COINS. 


307 


dwt.  grs. 

$  cts.  m. 

$  cts.  m. 

Ducat,  

2     5H 

2.15,4 

2.29,6 

Gold  Florin,  double  in  proportion, 

2     2 

1.57,6 

1.67,0 

HOLLAND. 

Double  Ryder,        - 

12  21 

11.44,2 

12.20,5 

6     9 

5.66,5 

6  04,3 

Ducat,   ------ 

2     5| 

2.13,3 

2.27,5 

Ten  Guilder  Piece,  Five  do.  in  pro., 

4     8 

3.78,0 

4.03,4 

MALTA. 

_ 

Double  Louis,        - 

10  16 

8.69,9 

9.27,8 

T 

50 

4.36  4 

4.85  2 

o 
2  16 

2.20^2 

2.  33  ',6 

MEXICO. 

Doubloon,  shares  in  proportion, 

17     9 

14.56,0 

15.53,5 

MILAN. 

Sequin,           - 

2     5| 

2.15,6 

2.29,0 

Doppia  or  Pistole,       - 

4    14 

3.57,2 

3.80.7 

Forty  Livre  Piece,  1808, 

8     8 

7.26,1 

7.74,2 

NAPLES. 

Six  Ducat  Piece,  1783. 

5  16 

4.92,5 

5  24,9 

Two  Ducat  Piece  or  Sequin,  1762, 

1  20| 

1.51,1 

1.59,1 

Three  Ducat  Piece  or  Oncetta,  1818, 

2  10* 

2.34,7 

2.49,0 

NETHERLANDS. 

Gold  Lion  or  14  Florin  Piece, 

5     7| 

4.73,1 

5.04,6 

Ten  Florin  Piece,  1820,      - 

4     7| 

3.76,6 

4.01,9 

PARMA. 

Quadruple  Pistole,  double  in  proportion, 

18     9 

15.59,6 

16.62,8 

Pistole  or  Doppia,  1787,     - 

4   14 

3.93,5 

4.19,4 

Pistole  or  Doppia,  1796, 

4  14 

3.87,5 

4.13,5 

Maria  Theresa,  1818, 

4     34 

3.62,4 

3.86,1 

PIEDMONT. 

Pistole,  coined  since  1785,  half  in  pro., 

5  20 

5.07,5 

5.41,1 

Sequin,  half  in  proportion, 

2     5 

2.13,7 

2.28,0 

Carlino,  coined  since  1785,  half  in  pro., 

29     6 

25.63,2 

27.34,0 

Piece  of  20  francs,  called  Marengo, 

4     34 

3.34,1 

3.56,4 

POLAND. 

2     5| 

2.13,7 

2.27,5 

PORTUGAL. 

Dobraon,        - 

34  12 

30.66,6 

32.70,6 

Dobra,       

18     6 

16.22,2 

17.30,1 

Johannes,       - 

18     0 

16.00,0 

17.06,4 

Moidore,  half  in  proportion, 

6  22 

6.14,9 

6.55,7 

Piece  of  16  Testoons  or  1600  Rees, 

2     6 

1.99,2 

2.12,1 

Old  Crusado  of  400  Rees,   - 

15 

.84,9 

.58,5 

New  Crusado  of  480  Rees,      - 

16£ 

.59,8 

.63,5 

Milree,  coined  in  1775, 

m 

.73,2 

.78,0 

308 


VALUE  OF  GOLD   COINS. 


[SECT.  LXXIV. 


PRUSSIA. 
Ducat,  1748, 

Ducat,  1787,      .... 
Frederick,  double,  1769, 

"        1800,     - 
"         single,  1778, 
"  "         1800,    - 

ROME. 

Sequin,  coined  since  1760, 
Scudo  of  Republic,     - 

RUSSIA. 

Ducat,  1796,          .... 
Ducat,  1763,      - 
Gold  Ruble,  1756, 
Gold  Ruble,  1799,      - 
Gold  Poltin,  1777,  ... 

Imperial,  1801,  - 
Half  Imperial,  1801,       ... 

SARDINIA. 
Carlino,  half  in  proportion, 

SAXONY. 

Ducat,  1784,          .... 
Ducat,  1797,      .... 
Augustus,  1754,     -        -  -  - 

Augustus,  1784,         ... 

SICILY. 

Ounce,  1751,         -         ... 
Double  Ounce,  1758,  - 

SPAIN. 

Doubloon,  1772,  parts  in  proportion, 
Doubloon,          -  - 

Pistole, 

Coronilla,  Gold  Dol.  or  Vintern,  1801, 

SWEDEN. 
Ducat, 

SWITZERLAND. 
Pistole  of  the  Helvetic  Republic,  1800, 

TREVES. 
Ducat,  -        -        - 

TURKEY. 

Sequin  Fonducli  of  Constan'ple,  1773, 
Sequin  Fonducli  of  Constan'ple,  1789, 
Half  Misseir,  1818, 
Sequin  Fonducli,         -         -         - 
Yeermeeblekblek,  .... 


dwt.  grs.    9  cts.  m. 

•  cts.  m. 

2  5| 

2.13,7 

2.27,9 

2  5| 

2.12,5 

2.26,7 

8  14 

7.47,5 

7.95,5 

8  14 

7.45,4 

7.95,1 

4  7 

3.74,9 

3.99,7 

4  7 

3.72,5 

3.97,5 

2  4| 

2.10,9 

2.25,1 

17  04 

14.82,8 

15.81,1 

2  6 

2.15,0 

2.29,7 

2  51 

2.12,5 

2.26,7 

1  04 

.90,9 

.96,7 

18| 

.69,1 

.73,7 

9 

.33,1 

.35,5 

7  174 

7.34,9 

7.82,9 

4  34 

3.68,9 

3.93,3 

10  74 

8.88,1 

9.47,2 

2  51 

2.12,5 

2.26,7 

2  5| 

2.13,7 

2.27,9 

4  64 

3.68,5 

3.92,5 

4  64 

3.72,5 

3  97,4 

2  204 

2.35,1 

2.50,4 

5  17 

4.72,7 

5.04,4 

17  84 

15.03,0 

16.02,8 

17  9 

14.56,0 

15.53,5 

4  84 

3.64,0 

3.88,4 

1  3 

.92,1 

.98,3 

2  5 

2.09,7 

2,23,5 

4  214 

4.27,9 

4.56,0 

2  5| 

2.02,5 

2.26,7 

2  51 

1.74,9 

1.86,8 

2  5| 

1.73,3 

1.84,8 

184 

.49,1 

.52,1 

2  5 

1.71,7 

1.83,0 

3  If 

2.84,0 

3.02,8 

SECT.  LXXV.] 


GEOMETRY. 


309 


TUSCANY. 

Zechino  or  Sequin,     - 
Ruspone  of  the  Kingdom  of  Etruria, 

VENICE. 
Zechino  or  Sequin,  shares  in  proportion, 

WURTEMBERG. 

Carolin,          ----- 
Ducat,       ----- 

ZURICH. 
Ducat,  double  and  half  in  proportion, 


d\vt.  grs. 

$  cts.  m. 

8  cts.  m. 

2     51 
6  17i 

2.16,6 
6.50,5 

2.31.8 
6.93,8 

2     6 

2.16,0 

2.31,0 

6     3i 
2     5 

4.59,4 
2.09,7 

4.89,8 
2.23,5 

2     5| 

2.12,5 

2.26,7 

SECTION  LXXV. 
GEOMETRY. 


DEFINITIONS. 

1.  A  point  is  that  which  has  position,  but  no  magnitude  nor 
dimensions  ;  neither  length,  breadth,  nor  thickness. 

2.  A  line  is  length,  without  breadth  or  thick- 
ness. 

3.  A  surface  or  superficies  is  an  extension, 
or  a  figure  of  two  dimensions,  length  and  breadth, 
but  without  thickness. 

4.  A  body  or  solid  is  a  figure  of  three  dimen- 
sions ;  viz.  length,  breadth,  and  thickness. 

5.  Lines  are  either  right  or  curved,  or  mixed 
of  these  two. 

6.  A  right  or  straight  line  lies  all  in  the  same 
direction   between   its   extremities,  and    is   the 
shortest  distance  between  two  points.     When  a 
tioned  simply,  it  means  a  right  line. 

7.  A  curve  continually  changes  its  direction  between  its  ex- 
treme points. 

8.  Lines  are  either  parallel,  oblique,  perpendicular,  or  tan- 
gential. 

9.  Parallel  lines  are  always  at  the  same  per-      

pendicular   distance,  and    never  meet,   though     • 

ever  so  far  produced. 


line 


is  men- 


310  GEOMETRY.  [SECT.  LXXT. 

10.  Oblique  lines  change  their  distance  from 
each  other,  and  would  meet  if  produced  on 
the  side  of  the  least  distance. 

11.  One  line  is  perpendicular  to  another,  when  it  inclines  not 
more  on  the  one  side  than  the  other,  or  when  the  angles  on  both 
sides  of  it  are  equal. 

12.  An  angle  is  the  inclination,  or  opening 
of  two  lines  having  different  directions,  and 
meeting  in  a  point. 

13.  Angles  are  right  or  oblique,  acute  or  obtuse. 

14.  A  right  angle  is  that  which  is  made  by 
one  line  perpendicular  to  another ;  or,  when 
the  angles  on  each  side  are  equal  to  one  an- 
other, they  are  right  angles. 

15.  An  oblique  angle  is  one  which  is  made 
by  two  oblique   lines,  and  is  either  less  or 
greater  than  a  right  angle. 

16.  An  acute  angle  is  less  than  a  right  angle. 

17.  An  obtuse  angle  is  greater  than  a  right  angle. 

18.  Superficies  are  either  plane  or  curved. 

19.  A  plane  superficies,  or  plane,  is  that  with  which  a  right 
line  may  every  way  coincide ;  or,  if  the  line  touch  the  plane  in 
two  points,  it  will  touch  it  in  every  point ;  but  if  not,  it  is  curved. 

20.  Plane  figures  are  bounded  either  by  right  lines  or  curves. 

21.  Plane  figures,  that  are  bounded  by  right  lines,  have  names 
according  to  the  number  of  their  sides,  or  of  their  angles ;  for 
they  have  as  many  sides  as  angles,  the  least  number  being  three. 

22.  A  figure  of  three  sides  and  angles  is  called  a  triangle ; 
and  it  receives  particular  denominations  from  the  relations  of 
its  sides  and  angles. 

23.  An  equilateral  triangle  is  that  whose 
three  sides  are  equal. 


24.  An  isosceles  triangle  is  that  which  has 
two  sides  equal. 

25.  A  scalene  triangle  is  that  whose  three 
sides  are  unequal. 

26.  A  right-angled  triangle  is  that  which 
has  one  right  angle. 


SECT.  LXXV.] 


GEOMETRY. 


311 


27.  Other  triangles  are  oblique-angled,  and  are  either  acute 
or  obtuse. 

28.  An  obtuse-angled  triangle  has  one  ob- 
tuse angle. 

29.  An  acute-angled  triangle  has  its  three  angles  acute. 

30.  A  figure  of  four  sides  and  angles  is  called  a  quadrangle, 
or  a  quadrilateral. 

31.  A  parallelogram  is  a  quadrilateral,  which  has  both  its 
pairs  of  opposite  sides  parallel  ;  and  it  takes  the  following  par- 
ticular names ;  viz.  rectangle,  square,  rhombus,  and  rhomboid. 

32.  A  rectangle  is  a  parallelogram,  having 
a  right  angle. 

33.  A  square  is  an  equilateral  rectangle, 
having  its  length  and  breadth  equal. 


34.  A  rhomboid  is  an  oblique-angled  par 
allelogram,  whose  opposite  sides  are  equal. 


35.  A  rhombus  is  a  parallelogram,  having 
all  its  sides  equal,  but  its  angles  oblique. 


36.  A  trapezium  is  a  quadrilateral,  which 
has  neither  two  of  its  opposite  sides  parallel. 

37.  A  trapezoid  has  only  one  pair  of  its 
opposite  sides  parallel. 

38.  A  diagonal  is  a  line  joining  any  two 
opposite  angles  of  a  quadrilateral. 

39.  Plane  figures  that  have  more  than  four  sides  are  in  gen- 
eral called  polygons  ;  and  they  receive  other  particular  names 
according  to  the  number  of  their  si'des  or  angles.     Thus, 

40.  A  pentagon  is  a  polygon  of  five  sides  ;  a  hexagon,  of  six 
sides  ;  a  heptagon,  of  seven  ;  an  octagon,  of  eight ;  a  nonagon, 
of  nine  ;  a  decagon,  of  ten  ;  an  undecagon,  of  eleven ;  and  a 
dodecagon,  of  twelve  sides. 

41.  A  regular  polygon  has  all  its  sides  and  all  its  angles 
equal.     If  they  are  not  both  equal,  the  polygon  is  irregular. 


312 


GEOMETRY. 


[SECT.  LXXT. 


II 


42.  An  equilateral  triangle  is  also  a  regular  figure  of  three 
sides,  and  the  square   is  one  of  four;  the  former  being  also 
called  a  trigon,  and  the  latter  a  tetragon. 

43.  Any  figure  is  equilateral  when  all  its  sides  are  equal ; 
and  it  is  equiangular  when  all  its  angles  are  equal.    When  both 
these  are  equal,  it  is  a  regular  figure. 

44.  A  circle  is  a  plane  figure,  bound- 
ed by  a  curve  line,  called  the  circum- 
ference, which    is    everywhere  equi- 
distant from  a  certain  point,  called  its 
centre.      The  circumference  itself  is 
often  called   a   circle,   and    also   the 
periphery. 

45.  The  radius  of  a  circle  is  a  line 
drawn  from  the  centre  to  the  circum- 
ference, as  A  B. 

46.  The  diameter  of  a  circle  is  a  line  drawn  through  the 
centre  and  terminating  at  the  circumference  on  both  sides,  as 
AC. 

47.  An  arc  of  a  circle  is  any  part  of  the  circumference,  as 
AD. 

48.  A  chord  is  a  right  line  joining  the  extremities  of  an  arch, 
asEF. 

49.  A  segment  is  any  part  of  a  circle  bounded  by  an  arc 
and  its  chord,  as  E  F  G. 

50.  A  semicircle  is  half  the  circle,  or  segment  cut  off  by  a 
diameter.      The  half  circumference  is  sometimes  called  the 
semicircle,  as  A  G  C. 

51.  A  sector  is  any  part  of  the  circle  bounded  by  an  arc  and 
two  radii  drawn  to  its  extremities,  as  A  B  H. 

52.  A  quadrant,  or  quarter  of  a  circle,  is  a  sector,  having  a 
quarter  of  its  circumference  for  its  arc,  and  its  two  radii  are 
perpendicular  to  each  other.     A  quarter  of  the  circumference 
is   sometimes    called   a   quadrant,   as 

ABD. 

53.  The  height  or  altitude  of  a  fig- 
ure is  a  perpendicular,  let  fall  from  an 
angle,  or  its  vertex,  to  the  opposite 
side,  called  the  base,  as  C  D. 

54.  In  a  right-angled  triangle   the 
side  opposite  to  the  right  angle  is  called 
the  hypothenuse,  and   the   other  two 
sides  are  called  the  legs,  and  some- 
times the  base  and  perpendicular ;  thus, 


D 


B 


SECT.  LXXV.]  GEOMETRY.  313 

A  B  is  the  base,  B  C  the  perpendicular,  and  A  C  the  hypothe- 
nuse. 

55.  When  an  angle  is  denoted  by 

three  letters,  of  which  one  stands  at  the  C  I) 

angular  point,  and  the  other  two  on  the  \          / 

two  sides,  that  which  stands  at  the  an-       p  \/  ^ 

gular  point  is  read  in  the  middle.  Thus,       "  ^ 

the  angle  contained  by  the  lines  B  A 

and  A  D  is  called  the  angle  B  A  D,  or  D  A  B. 

56.  The  circumference  of  every  circle  is  supposed  to  be  di- 
vided into  360  equal  parts,  called  degrees ;  and  each  degree  into 
60  minutes,  each  minute  into  60  seconds,  and  so  on.     Hence  a 
semicircle  contains  180  degrees,  and  a  quadrant  90  degrees. 

GEOMETRICAL  PROBLEMS. 

PROBLEM  I. 

To  divide  a  line,  A  B,  into  two  equal  parts. 

NOTE.  —  It  would  be  useful  for  the  pupil  to  be  furnished  with  a  pair  of 
dividers  and  a  rule,  and  to  be  required^to  draw  the  diagrams  here  given. 

Set  one  foot  of  the  dividers  in 
A,  and,  opening  them  beyond  the 
middle  of  the  line,  describe  arches 


B 


above  and  below  the  line  ;  with  the         ^ 

same  extent  of  the  dividers,  set  one  \     • 

foot  in  the  point  B,  and  describe  arch-  \  / 

es  crossing  the  former  ;  draw  a  line 

from  the  intersection  above  the  line 

to  the  intersection  below  the  line,  and  it  will  divide  the  line  AB 

into  two  equal  parts. 

PROBLEM  II. 

To  erect  a  perpendicular  on  the  point  C,  in  a  given  line. 

Set  one   foot  of  the    dividers   in  ^ 

the  given  point  C,  extend  the  other 
foot  to  any  distance  at  pleasure,  as 
to  D,  and  with  that  extent  make  the  M  /' 


marks  D  and  E.  With  the  dividers, 
one  foot  in  D,  at  any  extent  above 
half  the  distance  D  E,  describe  an 
arch  above  the  line,  and  with  the 
same  extent,  and  one  foot  in  E,  de- 
27 


\ 

E 


314  GEOMETRY.  [SECT.  LXXV. 

scribe  an  arch  crossing  the  former ;  draw  a  line  from  the  inter- 
section of  the  arches  to  the  given  point  C,  which  will  be  per- 
pendicular to  the  given  line  in  the  point  C. 

PROBLEM   III. 

To  erect  a  perpendicular  upon  the  end  of  a  line. 

Set  one  foot  of  the  dividers  in 
the  given  point  B,  open  them  to  any 
convenient  distance,  and  describe 
the  arch  C  D  E  ;  set  one  foot  in  C, 
and  with  the  same  extent  cross  the 


> 


*s 


arch  at  D ;   with  the  same  extent 

cross  the  arch  again  from  D  to  E  ; L 

then  with  one  foot  of  the  dividers  •"• 
in  D,  and,  with  any  extent  above  the  half  of  D  E,  describe  an 
arch  a  ;  take  the  dividers  from  D,  and,  keeping  them  at  the 
same  extent,  with  one  foot  in  E,  intersect  the  former  arch  a  in 
a  ;  from  thence  draw  a  line  to  the  point  B,  which  will  be  a 
perpendicular  to  A  B. 

PROBLEM  IV. 

From  a  given  point,  a,  to  let  fall  a  perpendicular  to  a  given  line 

AB. 

Set  one  foot  of  the  dividers  in 
the  point  a,  extend  the  other  so  as 
to  reach  beyond  the  line  A  B,  and 
describe  an  arch  to  cut  the  line  \ 


A  B  in  C  and  D  ;  put  one  foot  of        T     r?~~  — 
the  dividers  in  C,  and,  with  any 


..  -~r\ 


extent  above  half  C  D,  describe  an 
arch  b  ;  keeping  the  dividers  at  the  X. 

same  extent,  put  one  foot  in  D,  and 
intersect  the  arch  b  in  b ;    through 
which  intersection,  and  the  point  fl,  draw  aE,  the  perpendicu- 
lar required. 

PROBLEM  V. 

To  draw  a  line  parallel  to  a  given  line  A  B. 

Set  one  foot  of  the  dividers  in     E_ F 

any  part  of  the  line,  as  at  c;  ex- /'"' b~~^              /-'"a """ *N 
tend  the  dividers  at  pleasure,  un- 
less a  distance  be  assigned,  and  de-       A— B 

scribe  an  arch  £;  with  the  same 


SECT.  LXXV.] 


GEOMETRY. 


315 


extent  in  some  other  part  of  the  line  A  B,  as  at  e,  describe  the 
arch  a ;  lay  a  rule  to  the  extremities  of  the  arches,  and  draw 
the  line  E  F,  which  will  be  parallel  to  the  line  A  B. 

PROBLEM  VI. 

To  make  a  triangle  whose  sides  shall  be  equal  to  three  given 
lines,  any  two  of  which  are  longer  than  the  third. 

Let  ABC  be  the  three  given  lines;       ^ 

draw  a  line,  A  B,  at  pleasure  ;  take  the       r> 

line  C  in  the  dividers,  set  one  foot  in  A,  ^ 
and  with  the  other  make  a  mark  at  B ; 
then  take  the  given  line  B  in  the  dividers, 
and,  setting  one  foot  in  A,  draw  the  arch 
C  ;  then  take  the  line  A  in  the  dividers, 
and,  setting  one  foot  in  B,  intersect  the 
arch  C  in  C ;  lastly,  draw  the  lines  A  C 
and  B  C,  and  the  triangle  will  be  completed. 

PROBLEM  VII. 

To  make  a  square  whose  sides  shall  be  equal  to  a  given  line. 

Let  A  be  the  given  line  ;  draw  a 
line,  A  B,  equal  to  the  given  line  ;  from 
B  raise  a  perpendicular  to  C,  equal  to 
A  B  ;  with  the  same  extent,  set  one 
foot  in  C,  and  describe  the  arch  D  ; 
also,  with  the  same  extent,  set  one  foot 
in  A,  and  intersect  the  arch  D  ;  lastly, 
draw  the  line  A  D  and  C  D,  and  the 
square  will  be  completed. 

In  like  manner  may  a  parallelogram  be  constructed,  only  at- 
tending to  the  difference  between  the  length  and  breadth. 

PROBLEM    VIII. 

To  describe  a  circle,  which  shall  pass  through  any  three  given 

points,  not  in  a  straight  line. 
Let  the  three  given  points  be  A  B  C, 
through  which  the  circle  is  to  pass. 
Join  the  points  A  B  and  B  C  with  right 
lines,  and  bisect  these  lines ;  the  point 
D,  where  the  bisecting  lines  cross  each 
other,  will  be  the  centre  of  the  circle 
required.  Therefore,  place  one  foot 
of  the  dividers  in  D,  extending  the 
other  to  either  of  the  given  points,  and 


/ 

f->    ~ 

*-, 

/• 

A 

A 

316  GEOMETRY.  [SECT.  LXXT. 

the  circle  described  by  that  radius  will  pass  through  all  the 
points. 

Hence  it  will  be  easy  to  find  the  centre  of  any  given  circle  ; 
for,  if  any  three  points  are  taken  in  the  circumference  of  the 
given  circle,  the  centre  will  be  found  as  above.  The  same  may 
also  be  observed  when  only  a  part  of  the  circumference  is 
given. 

MENSURATION  OF  SOLIDS. 

DEFINITIONS. 

1.  Solids  are  figures  having  length,  breadth,  and  thickness. 


2.  A  prism  is  a  solid,  whose  ends  are 
any  plane  figures  which  are  equal  and 
similar,  and  its  sides  are  parallelograms. 

NOTE.  —  A  prism  is  called  a  triangular  prism, 
when  its  ends  are  triangles;  a  square  prism, 
when  its  ends  are  squares  ;  *  pentagonal  prism, 
when  its  ends  are  pentagons  ;  and  so  on. 


3.  A  cube  is  a  square  prism,  having 
six  sides,  which  are  all  squares. 


4.  A  parallelepiped  is  a  solid,  having 
six  rectangular  sides,  every  opposite  pair 
of  which  is  equal  and  parallel. 


5.  A  cylinder  is  a  round  prism,  having 
circles  for  its  ends. 


6.  A  pyramid  is  a  solid,  standing  on  a 
triangular,  square,  or  polygonal  basis,  and 
its  sides  are  triangles,  whose  vertices  meet 
in  a  point  at  the  top,  called  the  vertex  of 
the  pyramid. 


SECT.  LXXV.] 


GEOMETRY. 


317 


7.  A  cone  is  a  solid  figure,  having  a 
circle  for  its  base,  and  its  top  terminated 
in  a  point  or  vertex. 


8.  A  sphere  is  a  solid,  bounded  by  one 
continued  convex  surface,  every  point  of 
which    is    equally   distant    from  a  point 
within,  called  the  centre.      The  sphere 
may  be  conceived  to  be   formed  by  the 
revolution  of  a  semicircle  about  its  diam- 
eter, which  remains  fixed. 

A  hemisphere  is  half  a  sphere. 

9.  The  segment  of  a  pyramid,  sphere,  or  any  other  solid,  is  a 
part  cut  off  the  top  by  a  plane  parallel  to  the  base  of  that  figure. 

10.  A  frustum  is  the  part  that  remains  at  the  bottom  after 
the  segment  is  cut  off. 

11.  The  sector  of  a  sphere  is  composed  of  a  segment  less 
than  a  hemisphere,  and  of  a  cone  having  the  same  base  with 
the  segment,  and  its  vertex  in  the  centre  of  the  sphere. 

12.  The  axis  of  a  solid  is  a  tine  drawn  from  the  middle  of 
one  end  to  the  middle  of  the  opposite  end  ;  as  between  the  op- 
posite ends  of  a  prism.     The  axis  of  a  sphere  is  the  same  as  a 
diameter,  or  a  line  passing  through  the  centre,  and  terminating 
at  the  surface  on  both  sides. 

13.  The  height  or  altitude  of  a  solid  is  a  line  drawn  from 
its  vertex,  or  top,  perpendicular  to  its  base. 


MENSURATION  OF  SUPERFICIES   AND   SOLIDS. 

PROBLEM    I. 

To  find  the  area  of  a  square  or  parallelogram. 

RULE.  —  Multiply  the  length  by  the  breadth,  and  the  product  is  the 
superficial  contents. 

1.  What  are  the  contents  of  a  board  15  feet  long  and  2  feet 
wide  ?  Ans.  30  feet. 

2.  The  State  of  Massachusetts  is  about  128  miles  long  and 
48  miles  wide.     How  many  square  miles  does  it  contain  ? 

Ans.  6144  miles. 
27* 


318  GEOMETRY.  [SECT.  LXXY. 

3.  The  largest  of  the  Egyptian  pyramids  is  square  at  its  base, 
and  measures  693  feet  on  a  side.     How  much  ground  does  it 
cover  ?  Ans.  1 1  acres  4  poles. 

4.  What  is  the  difference  between  a  floor  40  feet  square,  and 
2  others,  each  20  feet  square  ?  Ans.  800  feet. 

5.  There  is  a  square  of  3600  yards  area  ;  what  is  the  side 
of  a  square,  and  the  breadth  of  a  walk  along  each  side  of  the 
square,  and  each  end,  which  may  take  up  just  one  half  of  the 
square  ?  A       $  42.42+  yards,  side  of  the  square. 

s-  I    8.78+  yards,  breadth  of  the  walk. 

PROBLEM    II. 

To  find  the  area  of  a  rhombus  or  rhomboid. 
RULE. —  Multiply  the  length  of  the  base  by  the  perpendicular  lieight. 

6.  The  base  of  a  rhombus  being  12  feet,  and  its  height  8  feet, 
required  the  area.  Ans.  96  feet. 

PROBLEM    III. 

To  find  the  area  of  a  triangle. 

RULE.  —  Multiply  the  base  by  half  the  perpendicular  height ;  or,  add 
the  three  sides  together;  then  take  half  of  that  sum,  and  out  of  it  sub- 
tract each  side  severally ;  multiply  the  half  of  the  sum  and  these  remain- 
ders together,  and  the  square  root  of  this  product  will  be  the  area  of  the 
triangle. 

7.  What  are  the  contents  of  a  triangle,  whose  perpendicular 
height  is  12  feet,  and  the  base  18  feet  ?  Ans.  108  feet. 

8.  There  is  a  triangle,  the  longest  side  of  which  is  15.6  feet, 
the  shortest  side  9.2  feet,  and  the  other  side  10.4  feet.     What 
are  the  contents  ?  Ans.  46. 139-|-  feet. 

PROBLEM  IV. 

Having  the  diameter  of  a  circle  given,  to  find  the  circumference. 

RULE.  —  Multiply  the  diameter  by  3.141592. 

NOTE.  —  The  exact  ratio  of  the  diameter  of  a  circle  to  its  circumference 
has  never  yet  been  ascertained.  Nor  can  a  square,  or  any  other  right- 
lined  figure,  be  found,  that  shall  be  exactly  equal  to  a  given  circle.  This 
is  the  famous  problem,  called  the  squaring  the  circle,  which  has  exercised 
the  abilities  of  the  greatest  mathematicians  for  ages,  and  has  been  the  oc- 
casion of  so  many  endless  disputes.  Van  Ceulen,  a  Dutchman,  was  the 
first  who  ascertained  this  ratio  to  any  great  degree  of  exactness,  which 
he  extended  to  thirty-six  places  of  decimals  ;  *  and  it  was  effected  by 


*  This  is  said  to  have  been  thought  so  curious  a  performance,  that  the 
numbers  were  cut  on  his  tombstone,  in  St.  Peter's  church-yard,  at  Leyden. 


SECT.  LXXV.]  GEOMETRY.  319 

means  of  the  continual  bisection  of  an  arc  of  a  circle.  This  process  was 
exceedingly  troublesome  and  laborious  ;  but  since  the  invention  of  Flux- 
ions  and  the  Summation  of  Infinite  Scries,  there  have  been  several  meth- 
ods found  for  doing  the  same  thing  with  less  labor  and  trouble,  and  far 
more  expedition.  If,  therefore,  the  diameter  of  a  circle  be  1  inch,  the 
circumference  will  be  3.1415926535897932384626433832795028841971693- 
9937510582097494459230781640628620899862803482534211706798214808- 
65132823066470938446460955051822317253594081284802  inches  nearly. 

9.  If  the  diameter  of  a  circle  is  144  feet,  what  is  the  circum- 
ference ?  Ans.  452.389248  feet. 

10.  If  the  diameter  of  the  earth  be  7964  miles,  what  is  its 
circumference  ?  Ans.  25019.638688+  miles. 

PROBLEM    V. 

Having  the  diameter  of  a  circle  given,  to  find  the  area. 

RULE.  —  Multiply  half  the  diameter  by  half  the  circumference,  and  the 
product  is  the  area ;  or,  which  is  the  same  thing,  multiply  the  square  of 
the  diameter  by  .785398,  and  the  product  is  the  area. 

Demonstration.  —  If  we  suppose  a  circle  to  be  divided  into 
an  infinite  number  of  triangles,  by  lines  drawn  from  the  centre 
of  the  circle  to  the  circumference,  we  may  find  the  contents  of 
each  triangle  by  multiplying  its  perpendicular  height  by  half  its 
base,  but  its  perpendicular  height  is  half  the  diameter  of  the 
circle,  and  half  its  base  is  half  a  certain  portion  of  the  circum- 
ference ;  and  all  the  bases  of  all  the  triangles  united  form  the 
whole  circumference. 

Again,  if  multiplying  half  the  circumference  by  half  the  di- 
ameter give  the  area  of  a  circle,  it  is  evident  that  the  area  will 
be  obtained  by  multiplying  one  fourth  the  circumference  by  the 
whole  diameter ;  and,  as  the  circumference  of  a  circle,  whose 
diameter  is  1,  is  3.141592,  therefore  by  multiplying  1  by  one 
fourth  of  3.141592  we  shall  obtain  the  area  of  a  circle  whose 
diameter  is  1.  Thus,  3.141592  X  i  —  .785398.  And  as  cir- 
cles are  to  each  other  as  the  squares  of  their  diameters  (see 
page  246),  therefore,  if  we  wish  to  obtain  the  area  of  a  circle 
whose  diameter  is  20  feet,  we  make  the  following  statement. 

As  I2  foot  :  202  feet  :  :  .785398  :  314.1592  feet,  Ans. 

And  this  process  is  equivalent  to  multiplying  the  square  of  the 
diameter  of  the  given  circle  by  .785398.  Q.  E.  D. 

11.  If  the   diameter  of  a  circle  be  761  feet,  what  is  the 
area  ?  Ans.  454840.475158  feet. 

12.  There  is  a  circular  island,  three  miles  in  diameter ;  how 
many  acres  does  it  contain  ?  Ans.  4523.89-f-  acres. 


320  GEOMETRY.  [SECT.  LXXV. 

PROBLEM  VI. 

Having  the  diameter  of  a  circle  given,  to  find  the  side  of  an 
equal  square.  • 

RULE.  —  Multiply  the  diameter  by  .886227,  and  the  product  is  the 
side  of  an  equal  square. 

Demonstration.  —  We  have  seen  in  Problem  V.  that  the  area 
of  a  circle,  whose  diameter  is  1,  is  .785398163397  ;  if,  there- 
fore,, we  extract  the  square  root  of  this  number,  we  shall  obtain 
the  side  of  a  square  of  a  circle  whose  diameter  is  1.  Thus, 
v/.785398163397:=  .886227.  And  since,  as  we  have  before 
stated,  the  diameters  of  circles  are  to  each  other  as  the  sides 
of  their  similar  inscribed  figures,  therefore,  as  1,  the  diame- 
ter of  the  given  circle,  is  to  the  diameter  of  the  required  circle, 
so  is  .886227,  the  side  of  a  square  equal  to  the  given  circle,  to 
the  side  of  a  square  equal  to  the  required  circle.  If,  therefore, 
the  diameter  of  a  circle  were  20  feet,  and  it  was  required  to 
find  the  side  of  a  square  that  would  contain  that  quantity,  we 
should  make  the  following  statement :  — 

As  1  foot  :  20  feet  :  :  .886227  :  17.72454  feet,  Ans. 

We  see,  from  this  process,  that  multiplying  the  diameter  of 
the  required  circle  by  .886227  gives  the  side  of  an  equal  square. 
Q.  E.  D. 

13.  I  have  a  round  field,  50  rods  in  diameter ;  what  is  the 
side  of  a  square  field,  that  shall  contain  the  same  area  ? 

Ans.  44.31135+  rods. 

PROBLEM    VII. 

Having  the  diameter  of  a  circle  given,  to  find  the  side  of  a 
square  inscribed. 

RULE.  —  Multiply  the  diameter  by  .707106,  and  the  product  is  the  side 
of  a  square  inscribed. 

Demonstration.  —  Let  A  B  C  D  represent 
a  square  inscribed  in  a  circle  whose  diam- 
eter is  1.  D  B  is  the  diameter  of  the  circle, 
and  it  is  also  the  diagonal  of  the  square. 
As  DAB  is  a  right-angled  triangle,  the 
squares  of  D  A  and  A  B  are  equal  to  the 
square  of  B  D  ;  but  AD  is  equal  to  A  B, 
therefore  the  square  of  D  A  is  equal  to  half 
the  square  of  B  D.  The  square  of  B  D  is  1,  therefore  the  square 
of  DA  is  .5  ;  and  the  square  root  of  .5  is  v.5  =  .707106  —  A  D, 
the  side  of  the  square,  whose  diameter  is  1.  Q.  E.  D.  There- 


SECT.  LXXV.]  GEOMETRY.  321 

fore,  to  find  the  side  of  a  square  inscribed  in  any  circle,  we 
say,  as  1  is  to  the  diameter  of  any  required  circle,  so  is  .707106 
to  the  side  of  a  square  inscribed  in  the  required  circle. 

14.  I  have  a  piece  of  timber  30  inches  in  diameter ;  how 
large  a  square  stick  can  be  hewn  from  it  ? 

Ans.  21.21-f-  inches  square. 

15.  Required  the  side  of  a  square,  that  may  be  inscribed  in 
a  circle  80  feet  in  diameter.  Ans.  56.56848-f-  feet. 

PROBLEM    VIII. 

In  a  given  circle  to  describe  a  hexagon  and  an  equilateral 
triangle,  and  to  find  the  length  of  one  of  the  sides  of  the  in- 
scribed triangle. 

RULE.  —  Multiply  the  diameter  by  .8660254  and  the  product  is  the 
side  of  an  inscribed  equilateral  triangle. 

Demonstration.  —  Let  A  B  C  D  E  F 
be  the  given  circle,  and  G  the  centre. 
From  the  point  B  in  the  circumference 
apply  the  radius  B  G  six  times  to  the 
circumference,  and  join  B  C,  CD,  D  E, 
EF,  FA,  and  AB,  and  the  figure 
A  B  C  D  E  F,  thus  formed,  is  an  equi- 
lateral  inscribed  hexagon. 

Join  the  alternate  angles  AE,  EC, 
and  C  A,  and  the  figure  A  E  C,  thus 
formed,  is  an  equilateral  triangle  inscribed.  It  is  equilateral 
because  the  three  sides  subtend  the  equal  arches  of  the  circum- 
ference. 

A  B  C  G  is  a  rhombus,  and  the  diagonal  B  G  is  equal  to  either 
side  of  the  rhombus.  If,  therefore,  the  diameter  of  the  circle 
A  D  is  1,  the  semidiameter  A  G  or  B  G  will  be  .5 ;  and  B  H, 
which  is  half  of  BG,  will  be  .25.  AHB  is  a  right-angled 
triangle,  and  therefore  A  H  is  equal  to  the  square  root  of  the 
difference  of  the  squares  of  A  B  and  B  H.  Thus  A  H  = 
-V/AB^^B  H2  =  A/&^£&  =  V.25  —  .0625  —  V.1875 
—  .4330127.  Now  if  AH  be  .4330127,  AC,  which  is  twice 
AH,  will  be  .8660254.  But  AC  is  the  side  of  the  equilateral 
triangle  inscribed  ;  and  as  we  have  before  shown,  page  246, 
that  all  similar  figures  are  in  proportion  to  their  homologous 
sides,  it  will  therefore  follow,  that  as  the  diameter  of  the  given 
circle,  which  is  1,  is  to  the  side  of  its  inscribed  equilateral  triangle 


322  •  GEOMETRY.  [SECT.  LXXV. 

.8660254,  so  is  the  diameter  of  any  other  circle  to  the  side  of 
its  inscribed  equilateral  triangle. 

16.  Required  the  side  of  an  equilateral  triangle,  that  may  be 
inscribed  in  a  circle  80  feet  in  diameter.     Ans.  69.28-|-  feet. 

17.  Required  the  side  of  an  equilateral  triangle,  that  may  be 
inscribed  in  a  circle  50  inches  in  diameter.      Ans.  43.3-f-in. 

18.  There  is  a  certain  piece  of  round  timber  30  inches  in  di- 
ameter ;  required  the  side  of  an  equilateral  triangular  beam, 
that  may  be  hewn  from  it.  Ans.  25.98-|-  inches. 

PROBLEM    IX. 

Having  the  circumference  of  a  circle  given,  to  find  the  di- 
ameter. 

RULE.  —  Multiply  the  circumference  by  .3183098,  and  the  product  is 
the  diameter. 

Rationale. —  As  we  have  seen  in  Problem  IV.,  the  ratio  of 
the  circumference  of  a  circle  to  its  diameter  is  as  3.141592  to 
1  ;  and  as  the  ratio  of  all  circumferences  of  circles  to  their  di- 
ameters is  the  same,  therefore,  if  the  circumference  of  a  circle 
be  1,  its  diameter  may  be  found  by  the  following  proportion  :  — 
As  3.141592 :!::!::  .3183098. 

Wherefore,  if  we  multiply  the  circumference  of  any  circle  by 
.3183098,  the  product  is  the  diameter.     Q.  E.  D. 

19.  If  the  circumference  of  a  circle  be  25,000  miles,  what 
is  its  diameter  ?  Ans.  7957.74-f-  miles. 

20.  If  the  circumference  of  a  round  stick  of  timber  be  50 
inches,  what  is  its  diameter  ?  Ans.  15.91549+  inches. 

PROBLEM   X. 

Having  the  circumference  of  a  circle  given,  to  find  the  side 
of  an  equal  square. 

RULE.  —  Multiply  the  circumference  by  .282094,  and  the  product  is 
the  side  of  an  equal  square. 

We  have  demonstrated  in  a  previous  problem,  that,  when  the 
diameter  of  a  circle  is  1,  the  side  of  an  equal  square  is  .886227  ; 
but  when  the  circumference  is  1,  the  side  of  an  equal  square 
must  be  as  much  less  than  this  number,  as  1  is  less  than 
3.141592;  that  is,  it  will  be  equal  to  the  number  of  times 
.886227  will  contain  3.141592 ;  .886227  -f-  3. 141592^.282094 ; 
therefore,  if  we  multiply  the  circumference  of  any  circle  by 
.282094,  the  product  is  the  side  of  an  equal  square.  Q.  E.  D. 


SECT.  LXXV.]  GEOMETRY.  323 

21.  I  have  a  circular  field  360  rods  in  circumference;  what 
must  be  the  side  of  a  square  field,  that  shall  contain  the  same 
quantity?  Ans.  101.55-|-  rods. 

22.  John  Smith  had  a  farm,  which  was  10,000  rods  in  cir- 
cumference, and  which  he  sold  at  $  71.75  per  acre,  and  he 
purchased  another  farm  containing  the  same  quantity  of  land 
in  the  form  of  a  square  ;  required  the  length  of  one  of  its  sides. 

Ans.  2820.94-f  rods. 

PROBLEM    XI. 

Having  the  circumference  of  a  circle  given,  to  find  the  side 
of  an  equilateral  triangle  inscribed. 

RULE.  —  Multiply  the  circumference  by  .2756646,  and  the  product  is 
the  side  of  an  equilateral  triangle  inscribed. 

Rationale.  —  We  have  seen  in  Problem  VIII.  that  the  ratio  of 
the  diameter  of  a  circle  to  its  inscribed  triangle  is  as  1  to 
.8660254 ;  but  the  ratio  of  the  circumference  of  a  circle  to  its 
diameter  is  as  3.141592  to  1 ;  therefore,  the  ratio  of  the  cir- 
cumference of  a  circle  to  its  inscribed  equilateral  triangle  is  as 
3.141592  to  .8660254  ;  that  is,  it  will  be  equal  to  the  number 
of  times  .8660254  will  contain  3.141592.  Thus  .8660254 -i- 
3.141592  =  .2756646.  Therefore,  by  multiplying  the  circum- 
ference of  any  circle  by  .2756646,  we  obtain  the  side  of  an 
equilateral  triangle  inscribed.  Q.  E.  D. 

23.  How  large  an  equilateral  triangle  may  be  inscribed  in  a 
circle,  whose  circumference  is  5000  feet  ?     Ans.  1378.323ft. 

24.  Required  the  side  of  an  equilateral  triangular  beam,  that 
may  be  hewn  from  a  round  piece  of  timber  80  inches  in  cir- 
cumference. Ans.  22.05-J-  inches. 

PROBLEM    XII. 

Having  the  circumference  of  a  circle  given,  to  find  the  side 
of  an  inscribed  square. 

RULE.  —  Multiply  the  circumference  by  .225079,  and  the  product  is 
the  side  of  a  square  inscribed. 

Rationale.  —We  have  seen  in  Problem  VII.  that  the  ratio  of 
the  diameter  of  a  circle  to  its  inscribed  square  is  as  1  to  .707106  ; 
we  have  also  seen  in  Problem  IV.  that  the  ratio  of  the  circum- 
ference of  a  circle  to  its  diameter  is  as  3.141592  to  1;  there- 
fore, the  ratio  of  the  circumference  of  a  circle  to  its  inscribed 
square  is  as  3.141592  to  .707106.  That  is,  it  will  be  equal  to 
the  number  of  times  .707106  will  contain  3.141592.  Thus 


g24  GEOMETRY.  [SECT.  LXXV. 

.707106  -T-  3.141592  =  .225079  ;  therefore,  if  the  circumfer- 
ence of  any  circle  be  multiplied  by  .225079,  the  product  is  the 
side  of  a  square  inscribed.  Q.  E.  D. 

25.  I  have  a  circular  field,  whose  circumference  is  5000  rods  ; 
what  is  the  side  of  a  square  field  that  may  be  made  in  it  ? 

Ans.  1125.395+  rods. 

26.  How  large  a  square  stick  may  be  hewn  from  a  piece  of 
round  timber  100  inches  in  circumference  ? 

Ans.  22.5+  inches  square. 

NOTE.  —  If  we  wish  to  find  the  circumference  of  a  tree,  which  will 
hew  any  given  number  of  inches  square,  we  divide  the  given  side  of  the 
square  by  .225079,  and  the  quotient  is  the  circumference  required. 

27.  What  must  be  the  circumference  of  a  tree  that  will  make 
a  beam  10  inches  square  ?  Ans.  44.42+  inches. 

28.  What  must  be  the  circumference  of  a  tree,  that,  when 
hewn,  it  may  be  18  inches  square  ?        Ans.  79.97+  inches. 

29.  I  have  a  garden  which  is  20  rods  square  ;  required  the 
circumference  of  a  circle,  in  feet,  that  will  inclose  this  garden. 

Ans.  1466.15+  feet 

PROBLEM     XIII. 

To  find  the  contents  of  a  cube  or  parallelopipedon. 

RULE.  —  Multiply  the  length,  height,  and  breadth  continually  to- 
gether, and  the  product  is  the  contents. 

30.  How  many  cubic  feet  are  there  in  a  cube,  whose  side  is 
18  inches  ?  Ans.  3|  feet. 

31.  What   are   the   contents  of  a   parallelopipedon,  whose 
length  is  6  feet,  breadth  2£  feet,  and  altitude  If  feet  ? 

Ans.  26£  feet. 

32.  How  many  cubic  feet  in  a  block  of  marble,  whose  length 
is  3  feet  2  inches,  breadth  2  feet  8  inches,  and  depth  2  feet  6 
inches  ?  Ans.  21    feet. 


PROBLEM    XIV. 

To  find  the  solidity  of  a  prism. 
RULE.  —  Multiply  the  area  of  the  base,  or  end,  by  the  height. 

33.  What  are   the   contents  of  a  triangular  prism,  whose 
length  is  12  feet,  and  each  side  of  its  base  24-  feet  ? 

Ans.  32.47+  feet. 

34.  Required  the  solidity  of  a  triangular  prism,  whose  length 
is  10  feet,  and  the  three  sides  of  its  triangular  end,  or  base,  are 
5,  4,  and  3  feet.  Ans.  60  feet. 


SECT.  LXXV.]  GEOMETRY.  325 

PROBLEM    XV. 

To  find  the  solidity  of  a  cone  or  pyramid. 
RULE.  —  Multiply  the  area  of  the  base  by  J  of  its  height. 

35.  What  is  the  solidity  of  a  cone,  whose  height  is  12£  feet, 
and  the  diameter  of  the  base  2£  feet  ?        Ans.  20.45-}-  feet. 

36.  What  are  the  contents  of  a  triangular  pyramid,  whose 
height  is  14  feet  6  inches,  and  the  sides  of  its  base  5,  6,  and  7 
feet  ?  Ans.  71.035+  feet. 

QUESTIONS  TO  EXERCISE  THE  ABOVE  PROBLEMS. 

37.  I  have  a  round  garden,  containing  75  square  rods ;  how 
large  a  square  garden  can  be  made  in  it  ? 

Ans.  47.7464-)-  square  rods. 

38.  I  have  a  circular  garden  containing  75  square  rods; 
what  must  be  the  side  of  a  square  field  that  would  contain  it  ? 

Ans.  9.772+  rods. 

39.  There  is  a  small  circular  field,  25  rods  in  diameter; 
what  is  the  difference  of  the  areas  of  the  inscribed  and  circum- 
scribed squares,  and  how  much  do  they  differ  from  the  areas  of 
the  field  ? 

Ans.  312.5  rods,  the  difference  of  the  squares;  134.12625+ 
rods,  the  circumscribed  square,  more  than  the  area ;  178.373 
rods,  inscribed  square  less  than  the  area. 

PROBLEM    XVI. 

To  find  the  surface  of  a  cone. 
RULE.  —  Multiply  the  circumference  of  the  base  by  half  its  slant  height. 

40.  What  is  the  convex  surface  of  a  cone,  whose  side  is  20 
feet,  and  the  circumference  of  its  base  9  feet  ?     Ans.  90  feet. 

PROBLEM     XVII. 

To  find  the  solidity  of  the  frustum  of  a  cone. 

RULE.  —  Multiply  the  diameters  of  the  two  bases  together,  and  to  thf 
product  add  $  of  the  square  of  the  difference  of  the  diameters ;  then  mul- 
tiply this  sum  by  .785398,  and  the  product  will  be  the  mean  area  between 
the  two  bases ;  lastly,  multiply  the  mean  area  by  the  length  of  the  frus- 
tum, and  the  product  will  be  the  solid  contents.  Or,  find  tlie  area  of  the 
two  ends  of  the  frustum,  multiply  those  two  areas  together,  and  extract 
tfte  square  root  of  their  product.  To  this  root  add  the  two  areas,  and 


326  GEOMETRY.  [SKCT.  LXXV. 

multiply  their  sum  by  one  third  of  the  altitude  or  length  of  the  frustum, 
and  the  product  will  be  the  solidity  of  the  frustum  of  a  cone,  or  pyramid. 

NOTE.  —  These  are  the  exact  rules  for  measuring  round  timber,  and 
should  be  adopted.  See  page  330. 

41.  What  are  the  contents  of  a  stick  of  timber,  whose  length 
is  40  feet,  the  diameter  of  the  larger  end  24  inches,  and  the 
smaller  end  12  inches  ?  Ans.  73£  feet,  nearly. 

PROBLEM     XVIII. 

To  find  the  solidity  of  a  sphere  or  globe. 
RULE.  —  Multiply  the  cube  of  the  diameter  by  .5236. 

42.  What  is  the  solidity  of  a  sphere,  whose  axis  or  diameter 
is  12  inches  ?  Ans.  904.78-j-  inches. 

43.  Required  the  contents  of  the  earth,  supposing  its  circum- 
ference to  be  25,000  miles. 

Ans.  263858149120.06886875  miles. 

PROBLEM    XIX. 

To  find  the  convex  surface  of  a  sphere,  or  globe. 
RULE.  —  Multiply  its  diameter  by  its  circumference. 

44.  Required  the  convex  surface  of  a  globe,  whose  diameter 
or  axis  is  24  inches.  Ans.  1809.55-f-  inches. 

45.  Required  the  surface  of  the  earth,  its  diameter  being 
7957£  miles,  and  its  circumference  25,000  miles. 

Ans.  198943750  square  miles. 

PROBLEM    XX. 

To  find  how  large  a  cube  may  be  cut  from  any  given  sphere, 
or  be  inscribed  in  it. 

RULE.  —  Square  the  diameter  of  the  sphere,  divide  that  product  by  3, 
and  extract  the  square  root  of  the  quotient  for  the  ansioer. 

Demonstration.  —  It  is  evident,  that,  if  a  cube  be  inscribed  in 
a  sphere,  its  corners  or  angles  will  be  in  contact  with  the  sur- 
face of  the  sphere,  and  that  a  line  passing  from  the  lower  cor- 
ner of  the  cube  to  its  opposite  upper  corner  will  be  the  diame- 
ter of  the  sphere  ;  and  that  the  square  of  this  oblique  line  is 
equal  to  the  sum  of  the  squares  of  three  sides  of  the  inscribed 
cube  is  evident,  from  the  fact  that  the  square  of  any  two  sides 
of  the  cube  (suppose  two  sides  at  the  base)  is  equal  to  the 
square  of  the  diagonal  across  the  base  ;  and  that  the  square  of 


SECT.  LXXVI.]  GAUGING.  327 

this  diagonal  (which  we  have  just  proved  to  be  equal  to  the 
square  of  two  sides  at  the  base)  and  the  square  of  the  height 
of  the  cube  are  equal  to  the  square  of  the  diagonal  line,  which 
passes  from  the  lower  corner  of  the  square  to  the  opposite  up- 
per corner,  which  line  is  the  diameter  of  the  sphere.  There- 
fore, the  square  of  the  diameter  of  any  sphere  is  equal  to  the 
sum  of  the  squares  of  any  three  sides  of  an  inscribed  cube ;  or 
•J  of  the  square  of  the  diameter  of  any  sphere  is  equal  to  the 
square  of  one  of  the  sides  of  an  inscribed  cube.  Q.  E.  D. 

46.  How  large  a  cube  may  be  inscribed  in  a  globe  12  inches 
in  diameter  ?  Ans.  6.928-j-in.  in  the  side  of  the  cube. 

12  v  12 

-  —  48 ;  v48  =  6.928+  Answer. 
o 

47.  How  large  a  cube  may  be  inscribed  in  a  sphere  40  inch- 
es in  diameter  ?  Ans.  23.09-f-  inches. 

48.  How  many  cubic  inches  are  contained  in  a  cube,  that 
may  be  inscribed  in  a  sphere  20  inches  in  diameter  ? 

Ans.  1539.6+  inches. 


SECTION  LXXVI. 
GAUGING. 

GAUGING  is  the  art  of  finding  the  contents  of  any  regular  ves- 
sel, in  gallons,  bushels,  &c. 

PROBLEM    I. 

To  find  the  number  of  gallons,  &c.,  in  a  square  vessel. 

RULE. —  Take  the  dimensions  in  inches;  then  multiply  the  length, 
breadth,  and  height  together ;  divide  the  product  by  282  for  ale  gallons, 
231  for  wine  gallons,  and  2 150.42  for  bushels. 

1.  How  many  wine  gallons  will  a  cubical  box  contain,  that 
is  10  feet  long,  5  feet  wide,  and  4  feet  high  ? 

Ans.  1496T8Tgal. 

2.  How  many  ale  gallons  will  a  trough  contain,  that  is  12 
feet  long,  6  feet  wide,  and  2  feet  high  ?  Ans.  882|f  gal. 

3.  How  many  bushels  of  grain  will  a  box  contain,  that  is 
15  feet  long,  5  feet  wide,  and  7  feet  high  ?        Ans.  421.8bu. 


328  TONNAGE  OF  VESSELS.  [SECT.  LXXVII. 

PROBLEM    II. 

To  find  the  contents  of  a  cask. 

RULE. —  Take  the  dimensions  of  live  cask  in  inches  ;  namely,  the  dl 
ameter  of  the  bung  and  head,  and  the  length  of  the  cask.  Note  the  dif- 
ference between  the  bung  diameter  and  t/te  head  diameter. 

If  the  staves  of  the  cask  be  much  curved  between  the  bung  and  the  head, 
multiply  the  difference  by  .7  ;  if  not  quite  so  much  curved ,  by  .65  ;  if 
they  bulge  yet  less,  by  .6  ;  and  if  they  are  almost  straight,  by  .55  ;  add 
the  product  to  the  head  diameter ;  the  sum  will  be  a  mean  diameter  by 
which  the  cask  is  reduced  to  a  cylinder. 

Square  the  mean  diameter  thus  found,  then  multiply  it  by  the  length  ; 
divide  the  product  by  359  for  ale  or  beer  gallons,  and  by  294  for  wine 
gallons. 

4.  Required  the  contents  in  wine  gallons  of  a  cask,  whose 
bung  diameter  is  35  inches,  head  diameter  27  inches,  and  length 
45  inches. 

35  __  27  x  .7  =    5.6       32.6  X  32.6  x  45  =  47824.20 
27  +  5.6         =32.6          478242Q 

— 294 —  —  162.66  wine  gallons. 

5.  What  are  the  contents  of  a  cask  in  ale  gallons,  whose 
bung  diameter  is  40  inches,  head  diameter  30  inches,  and  length 
50  inches  ?  Ans.  185.55+  ale  gallons. 

PROBLEM    III. 

To  find  the  contents  of  a  round  vessel,  wider  at  one  end  than 
the  other. 

RULE.  — Multiply  the  greater  diameter  by  the  less;  to  this  product 
add  £  of  the  square  of  their  difference,  then  multiply  by  the  height,  and 
divide  as  in  the  last  rule. 

6.  What  are  the  contents  in  wine  measure  of  a  tub,  40 
inches  in  diameter  at  the  top,  30  inches  at  the  bottom,  and 
whose  height  is  50  inches  ?  Ans.  209.75  wine  gallons. 


SECTION  LXXVII. 
TONNAGE   OF  VESSELS. 

CARPENTER'S  RULE.  —  For  single-decked  vessels,  multiply  the 
length,  breadth  at  the  main  beam,  and  depth  in  the  hold  together,  and 
divide  the  product  by  95,  and  the  quotient  is  the  tons 


SECT.  LXXVII.]  TONNAGE  OF  VESSELS.  329 

But  for  a  double-decked  vessel,  take  half  of  the  breadth  of  the  main 
beam  for  the  depth  of  the  hold,  and  proceed  as  before. 

1.  What  is  the  tonnage  of  a  single-decked  vessel,  whose  length 
is  65ft.,  breadth  20ft.,  and  depth  10ft.  ?         Ans.  136f|  tons. 

2.  What  is  the  tonnage  of  a  double-decked  vessel,  whose 
length  is  70  feet,  and  breadth  24  feet  ?         Ans.  212T4g-  tons. 

GOVERNMENT  RULE.  —  If  the  vessel  be  double-decked,  take  the 
length  thereof  from  the  fore  part  of  the  main  stem  to  the  after  part  of 
the  stern-post  above  the  upper  deck  ;  the  breadth  thereof  at  the  broad- 
est part  above  the  main  wales,  half  of  which  breadth  shall  be  accounted 
the  depth  of  such  vessel,  and  then  deduct  from  the  length  %  of  the 
breadth ;  multiply  the  remainder  by  the  breadth ,  and  the  product  by 
the  depth,  and  divide  this  last  product  by  95,  the  quotient  whereof 
shall  be  deemed  the  true  contents  or  tonnage  of  such  ship  or  vessel ; 
and  if  such  ship  or  vessel  be  single-decked,  take  the  length  and  breadth, 
as  above  directed,  deduct  from  the  said  length  f  of  the  breadth,  and 
take  the  depth  from  the  under  side  of  the  deck-plank  to  the  ceiling  in 
the  hold,  and  then  multiply  and  divide  as  aforesaid,  and  the  quotient 
shall  be  deemed  the  tonnage. 

3.  What  is  the  government  tonnage  of  a  single-decked  ves- 
sel, whose  length  is  70  feet,  breadth  30  feet,  and  depth  in  the 
hold  9  feet  ?  Ans.  147|f-  tons. 

4.  What  is  the  government  tonnage  of  a  single-decked  ves- 
sel, whose  length  is  75  feet,  breadth  22  feet,  and  depth  in  the 
hold  12  feet?  Ans.  171ff£  tons. 

5.  What  is  the  government  tonnage  of  a  double-decked  ves- 
sel, which  has  the  following  dimensions  :  length  98  feet,  breadth 
35  feet  ?  Ans.  496££  tons. 

6.  Required  the  government  tonnage  of  a  double-decked  ves- 
sel, whose  length  is  180  feet,  and  breadth  40  feet. 

Aris.  1313|f  tons. 

7.  Required  the  government  tonnage  of  a  single -decked  ves- 
sel, whose  length  is  78  feet,  width  21  feet,  and  depth  9  feet. 

Ans.  130¥5T\  tons. 

8.  What  is  the  government  tonnage  of  a  double-decked  vessel, 
whose  length  is  159ft.,  and  width  30ft.  ?        Ans.  667 J-£  tons. 

9.  What  is  the  government  tonnage  of  Noah's  ark,  admitting 
its  length  to  have  been  479  feet,  its  breadth  80  feet,  and  its 
depth  48  feet.  Ans.  1742 1T9^  tons. 

10.  What  is  the  government  tonnage  of  a  vessel,  whose  length 
is  200  feet,  and  breadth  35  feet  ?  Ans.  1153/F  tons. 

11.  The  new  ship  Montezuma  is  280  feet  in  length,  and  40 
feet  in  breadth.     Required  the  government  tonnage. 

Ans.  2155-U  tons. 
28* 


330  MENSURATION  OF  LUMBER.      [SECT.  LXXVIII. 

SECTION  LXXVIII. 
MENSURATION  OF  LUMBER. 

PROBLEM    I. 

To  find  the  contents  of  a  board. 

RULE.  —  Multiply  the  length  of  the  board,  taken  in  feet,  by  its  breadth 
taken  in  inches,  divide  this  product  by  12,  and  the  quotient  is  the  con- 
tents in  square  feet. 

1.  What  are  the  contents  of  a  board  24  feet  long,  and  8 
inches  wide  ?  Ans.  16  feet. 

2.  What  are  the  contents  of  a  board  30  feet  long,  and  16 
inches  wide  ?  Ans.  40  feet. 

PROBLEM    II. 

» 

To  find  the  contents  of  joists. 

RULE.  —  Multiply  the  depth  and  width,  taken  in  inches,  by  the  length 
in  feet;  divide  this  product  by  12,  and  the  quotient  is  the  contents  inject. 

3.  How  many  feet  are  there  in  3  joisis,  which  are  15  feet 
long,  5  inches  wide,  and  3  inches  thick  ?  Ans.  56£  feet. 

4.  How  many  feet  in  20  joists,  10  feet  long,  6  inches  wide, 
and  2  inches  thick  ?  Ans.  200  feet. 

PROBLEM    III. 

To  measure  round  timber. 

We  have  inserted  below  the  rule  usually  adopted  by  survey- 
ors of  lumber  ;  but  it  is  a  very  unjust  rule,  if  it  is  intended  to 
give  only  40  cubic  feet  of  timber  for  a  ton.  For,  if  a  stick  of 
round  timber  be  40  feet  long,  and  its  circumference  be  48 
inches,  it  is  considered  by  surveyors  to  contain  one  ton,  or  40 
feet ;  whereas,  it  in  reality  contains,  according  to  the  following 
correct  process,  50^  cubic  feet. 

OPERATION. 

48  x  .31831=15.27888 ;  15.27888  -^  2  =  7.63944 ;  7.63944 
X  24=  183.34656 ;  183.34656  X  40  =  7333.8624  ;  7333.8624 
-f-  144  =  50T9^-,  that  is,  it  contains  as  many  cubic  feet  as  a 
stick  SOy9^  feet  long  and  1  foot  square. 

RULE.  —  Multiply  the  length  of  the  stick,  taken  in  feet,  by  the  square 


SECT.  LXXIX.]         PHILOSOPHICAL  PROBLEMS.  331 

of±  the  girth,  taken  in  inches  ;  divide  this  product  by  144,  and  the  quo- 
tient is  the  contents  in  cubic  feet. 

NOTE.  —  The  girth  is  usually  taken  about  £  of  the  distance  from  the 
larger  to  the  smaller  end. 

5.  How  many  cubic  feet  in  a  stick  of  timber,  which  is  30 
feet  long,  and  whose  girth  is  40  inches  ?  Ans.  20£  feet. 

6.  If  a  stick  of  timber  is  50  feet  long,  and  its  girth  is  56  inches, 
what  number  of  cubic  feet  does  it  contain  ?     Ans.  GSyV  feet. 

7.  What  are  the  contents  of  a  log  90  feet  long,  and  whose 
circumference  is  120  inches  ?  Ans.  562£  feet. 


SECTION  LXXIX. 
PHILOSOPHICAL   PROBLEMS.* 

PROBLEM    I. 

To  find  the  time  in  which  pendulums  of  different  lengths 
would  vibrate  ;  that  which  vibrates  seconds  being  39.2  inches. 

The  time  of  the  vibrations  of  pendulums  are  to  each  other 
as  the  square  roots  of  their  lengths ;  or  their  lengths  are  as  the 
squares  of  their  times  of  vibrations. 

RULE.  — As  the  square  of  one  second  is  to  the  square  of  the  time  in 
seconds  in  which  a  pendulum  would  vibrate,  so  is  39.2  inches  to  the 
length  of  the. required  pendulum. 

EXAMPLES. 

1.  Required  the  length  of  a  pendulum  that  vibrates  once  in 
8  seconds. 

I2  :  82  :  :  39.2in.  :  2508.8in.  =  209T^  feet,  Ans. 

2.  Required  the  length  of  a  pendulum   that  shall  vibrate  4 
times  a  second.  Ans.  2^y  inches. 

3.  Required  the  length  of  a  pendulum  that  shall  vibrate  once 
a  minute.  Ans.  3920  yards. 

4.  How  often  will  a  pendulum  vibrate  whose  length  is  100 
feet  ?  Ans.  once  in  5.53+  seconds. 

PROBLEM    II. 

To  find  the  weight  of  any  body,  at  any  assignable  distance 
above  the  earth's  surface. 

*  For  demonstrations  of  the  following  problems,  the  student  is  referred 
to  Enfield's  Philosophy,  or  to  the  Cambridge  Mathematics. 


332  PHILOSOPHICAL  PROBLEMS.         [SECT.  LXXIX. 

The  gravity  of  any  body  above  the  earth's  surface  decreases, 
as  the  squares  of  its  distance  in  semidiameters  of  the  earth 
from  its  centre  increases. 

RULE.  —  As  the  square  of  the  distance  from  the  earth's  centre  is  to 
the  square  of  the  earth's  semidiameter ,  so  is  the  weight  of  the  body  on 
the  earth's  surface  to  its  weight  at  any  assignable  distance  above  the 
surface  of  the  earth,  and  vice  versa. 

5.  If  a  body  weigh  900  pounds  at  the  earth's  surface,  what 
would  it  weigh  2000  miles  above  its  surface  ?      Ans.  4001bs. 

6.  Admitting  the  semidiameter  of  the  earth  to  be  4000  miles, 
what  would  be  the  weight  of  a  body  20,000  miles  above  its  sur- 
face, that  on  its  surface  weighed  144  pounds?         Ans.  41bs. 

7.  How  far  must  a  body  be  raised  to  lose  half  its  weight  ? 

Ans.  1656.85-+-  miles. 

8.  If  a  man  at  the  earth's  surface  could  carry  150  pounds, 
how  much  would  that  burden  weigh  at  die  earth,  which   he 
could  sustain  at  the  distance  of  the  moon,   whose  centre  is 
240,000  miles  from  the  earth's  centre  ?         Ans.  540,0001bs. 

9.  If  a  body  at  the  surface  of  the  earth  weigh  900  pounds, 
but  being  carried  to  a  certain  height  weighs  only  400  pounds, 
what  is  that  height  ?  Ans.  2000  miles. 

PROBLEM    III. 

By  having  the  height  of  a  tide  on  the  earth  given,  to  find  the 
height  of  one  at  the  moon. 

RULE.  —  As  the  cube,  of  the  moon's  diameter,  multiplied  by  its  density, 
is  to  the  cube  of  the  earth's  diameter,  multiplied  by  its  density,  so  is  the 
height  of  a  tide  on  the  earth  to  the  Jieight  of  one  at  the  moon. 

10.  The  moon's  diameter  is  2180  miles,  and  its  density  494  ; 
the  earth's  diameter  is  7964  miles,  and  its  density  400.     If, 
then,  by  the  attraction  of  the  moon,  a  tide  of  6  feet  is  raised  at 
the  earth,  what  will  be  the  height  of  a  tide  raised  by  the  attrac- 
tion of  the  earth  at  the  moon  ?  Ans.  236.8-f-  feet. 

NOTE.  —  The  above  question  is  on  the  supposition  that  the  moon  has 
seas  and  oceans  similar  to  those  on  the  earth,  but  astronomers  at  the  pres- 
ent time  doubt  their  existence  at  this  secondary  planet. 

PROBLEM   IV. 

To  find  the  weight  of  a  body  at  the  sun  and  planets,  having 
its  weight  given  at  the  earth. 

If  the  diameters  of  two  globes  be  equal,  and  their  densities 
different,  the  weight  of  a  body  on  their  surfaces  will  be  as  their 
densities. 


SECT.  LXXIX.]         PHILOSOPHICAL  PROBLEMS. 


333 


If  their  densities  be  equal,  and  their  diameters  different,  the 
weight  of  a  body  will  be  as  ^  of  their  circumference. 

If  their  diameters  and  densities  be  both  different,  the  weight 
of  a  body  will  be  as  f  of  their  semidiameters,  multiplied  by 
their  densities. 

Therefore,  having  the  weight  of  a  body  on  the  surface  of  the 
earth  given,  to  find  its  weight  at  the  surface  of  the  sun  and  the 
several  planets,  we  adopt  the  following 

RULE.  —  As  §  of  the  earth's  semidiameter,  multiplied  by  its  density, 
is  to  $  of  the  sun's  or  planet's  semidiameter,  multiplied  by  its  density, 
so  is  the  weight  of  a  body  at  the  surface  of  tlie  earth  to  the  weight  of  a 
body  at  the  surface  of  the  sun  or  planet. 

11.  If  the  weight  of  a  man  at  the  surface  of  the  earth  be  170 
pounds,  what  will  be  his  weight  at  the  surface  of  the  sun,  and 
the  several  planets,  whose  densities,  &c.,  are  as  in  the  following 
table  ? 


Sun, 
Jupiter, 
Saturn, 
Earth, 
Moon, 

Density. 

Diameter. 

Semidiameter. 

§  Semidiameter. 

100 
94.5 
67 
400 
494 

883246 
89170 
79042 
7964 
2180 

441623 
44585 
39521 
3982 
1090 

294415 
29723 
26347 
2654 
726 

Ans.  Weight  at  the  sun  4714.6-f-lbs.,  at  Jupiter  449.7-j-lbs., 
at  Saturn  282.6-j-lbs.,  at  the  moon  57.4+lbs. 

PROBLEM   V. 

To  find  how  far  a  heavy  body  will  fall  in  a  given  time,  near 
the  surface  of  the  earth. 

Heavy  bodies  near  the  surface  of  the  earth  fall  16  feet  in  one 
second  of  time  ;  and  the  velocities  they  acquire  in  falling  are 
as  the  squares  of  the  times ;  therefore,  to  find  the  distance  any 
body  will  fall  in  a  given  time,  we  adopt  the  following 

RULE.  — As  the  square  of  I  second  is  to  the  square  of  the  time  in 
seconds  that  the  body  is  falling,  so  is  16  feet  to  the  distance  in  feet  that 
the  body  will  fall  in  the  given  time. 

12.  How  far  will  a  leaden  bullet  fall  in  8  seconds  ? 

I2  :  82  :  :  16ft.  :  1024ft.  =  Answer. 

13.  How  far  would  a  body  fall  in  1  minute  ? 

Ans.  10  miles  1600  yards. 


334  PHILOSOPHICAL  PROBLEMS.         [SECT.  LXXIX. 

14.  How  far  would  a  body  fall  in  1  hour  ? 

Ans.  39,272  miles  1280  yards. 

15.  How  far  would  a  body  fall  in  9  days  ? 

Ans.  1,832,308,363  miles  1120  yards. 

PROBLEM    VI. 

The  velocity  given,  to  find  the  space  fallen  through  to  ac- 
quire that  velocity. 

RULE.  —  Divide  the  velocity  by  8,  and  the  square  of  the  quotient  will 
be  the  distance  fallen  through  to  acquire  that  velocity. 

16.  The  velocity  of  a  cannon-ball  is  660  feet  per  second. 
From  what  height  must  it  fall  to  acquire  that  velocity  ? 

Ans.  6806±  feet. 

17.  At  what  distance  must  a  body  have  fallen  to  acquire  the 
velocity  of  1000  feet  per  second  ?      Ans.  2  miles  5065  feet. 

PROBLEM  VII. 

The  velocity  given  per  second,  to  find  the  time. 

RULE.  —  Divide  tJte  velocity  by  8,  and  a  fourth  part  of  tlte  quotient 
will  be  the  time  in  seconds. 

18.  How  long  must  a  body  be  falling  to  acquire  a  velocity 
of  200  feet  per  second  ?  Ans.  6£  seconds. 

19.  How  long  must  a  body  be  falling  to  acquire  a  velocity  of 
320  feet  per  second  ?  AJIS.  10  seconds. 

PROBLEM   VIII. 

The  space  through  which  a  body  has  fallen  given,  to  find  the 
time  it  has  been  falling. 

RULE.  —  Divide  the  square  root  oftJie  space  in  feet  fallen  through  by 
4,  and  the  quotient  will  be  the  time  in  seconds  in  which  it  was  falling. 

20.  How  long  would  a  body  be  falling  through  the  space  of 
40,000  feet  ?  Ans.  50  seconds. 

21.  How  long  would  a  ball  be  falling  from  the  top  of  a  tow- 
er, that  was  400  feet  high,  to  the  earth  ?         Ans.  5  seconds. 

PROBLEM    IX. 

The  weight  of  a  body  and  the  space  fallen  through  given,  to 
find  the  force  with  which  it  will  strike. 

RULE.  —  Multiply  the  space  fallen  through  by  64,  then  multiply  the 
square  root  of  this  product  by  the  weight,  and  the  product  is  the  momen- 
tum^ or  force  with  which  it  will  strike. 


SECT.  LXXX.]  MECHANICAL   POWE 


22.  If  the  rammer  for  driving  the   pi  IPS  of  Warren  Bridge 
weighed  1000  pounds,  and  fell  through  a  spa^^^J^qt,  tyfthi 
what  force  did  it  strike  the  pile  ?  :^^:=^— :. 

Vl6  X  64  =  32  32  X  1000  =  32,0001bs.  Answer. 

23.  Bunker  Hill  Monument  is  220  feet  in  height;    what 
would  be  the  momentum  of  a  stone,  weighing  4  tons,  falling 
from  the  top  to  the  ground  ?  Ans.  255,7761bs. 


SECTION  LXXX. 
MECHANICAL    POWERS. 

THAT  body  which  communicates  motion  to  another  is  called 
the  power. 

The  body  which  receives  motion  from  another  is  called  the 
weight. 

The  mechanical  powers  are  six,  the  Lever,  the  Wheel  and 
Axle,  the  Pulley,  the  Inclined  Plane,  the  Screw,  and  the  Wedge. 

THE   LEVER. 

The  lever  is  a  bar,  movable 
about  a  fixed  point,  called  its 
fulcrum  or  prop.  It  is  in  theory 
considered  as  an  inflexible  line, 
without  weight.  It  is  of  three 
kinds  ;  the  first,  when  the  ^>rop 

is  between  the  weight  and  the  power  ;  the  second,  when  the 
weight  is  between  the  prop  and  the  power ;  the  third,  when  the 
power  is  between  the  prop  and  the  weight. 

A  power  and  weight  acting  upon  the  arms  of  a  lever  will 
balance  each  other,  when  the  distance  of  the  point  at  which  the 
power  is  applied  to  the  lever  from  the  prop  is  to  the  distance 
of  the  point  at  which  the  weight  is  applied  as  the  weight  is  to 
the  power. 

Therefore,  to  find  what  weight  may  be  raised  by  a  given 
power,  we  adopt  the  following 

RULE.  —  As  the  distance  between  the  body  to  be  raised,  or  balanced, 
and  the  fulcrum  or  prop,  is  to  the  distance  between  the  prop  and  the 
point  where  the  power  is  applied,  so  is  the  power  to  the  weight  which  it 
will  balance. 


336 


MECHANICAL  POWERS. 


[SECT.  LXXX. 


1.  If  a  man  weighing  170  pounds  be  resting  upon  a  lever  10 
feet  long,  what  weight  will  he  balance  on  the  other  end,  the 
prop  being  one  foot  from  the  weight  ?  Ans.  15301bs. 

2.  If  a  weight  of  1530  pounds  were  to  be  raised  by  a  lever 
10  feet  long,  and  the  prop  fixed  one  foot  from  the  weight,  what 
power  applied  to  the  other  end  of  the  lever  would  balance  it  ? 

Ans.  1701bs. 

3.  If  a  weight  of  1530  pounds  be  placed  one  foot  from  the 
prop,  at  what  distance   from  the  prop  must  a  power  of  170 
pounds  be  applied  to  balance  it  ?  Ans.  9  feet. 

4.  At  what  distance  from  a  weight  of  1530  pounds  must  a 
prop  be   placed,  so  that  a  power  of  170  pounds,  applied  9 
feet  from  the  prop,  may  balance  it  ?  Ans.  1  foot. 

5.  Supposing  the  earth  to  contain  4,000,000,000,000,000,000,- 
000  cubic  feet,  each  foot  weighing  100  pounds,  and  that  the 
earth  was  suspended  at  one  end  of  a  lever,  its  centre  being  6000 
miles  from  the  fulcrum  or  prop,  and  that  a  man  at  the  other 
end  of  the  lever  was  able  to  pull,  or  press  with  a  force  of  200 
pounds ;  what  must  be  the  distance  between  the  man  and  the 
fulcrum,  that  he  might  be  able  to  move  the  earth  1 

Ans.  12,000,000,000,000,000,000,000,000  miles. 

6.  Supposing  the  man  in  the  last  question  to  be  able  to  move 
his  end  of  the  lever  100  feet  per  second,  how  long  would  it 
take  him  to  raise  the  earth  one  inch  ? 

Ans.  52,813,479,690y.  17d.  14h.  57m.  46§sec. 

THE   WHEEL   AND  AXLE. 

The  wheel  and  axle  is  a  wheel 
turning  round  together  with  its 
axle ;  the  power  is  applied  to  the 
circumference  of  the  wheel,  and 
the  weight  to  that  of  the  axle  by 
means  of  cords. 

An  equilibrium  is  produced  in 
the  wheel  and  axle,  when  the 
weight  is  to  the  power  as  the  di- 
ameter of  the  wheel  to  the  diam- 
eter of  its  axle. 

To  find,  therefore,  how  large  a 
power  must  be  applied  to  the 
wheel  to  raise  a  given  weight  on 
the  axle,  we  adopt  the  following 


SECT.    LXXX.] 


MECHANICAL   POWERS. 


337 


RULE.  — ••  As  the  diameter  of  the  wheel  is  to  the  diameter  of  the  axle, 
so  is  the  weight  to  be  raised  by  the  axle  to  the  power  that  must  be  applied 
to  the  wheel. 

7.  If  the  diameter  of  the  axle  be  6  inches,  and  the  diameter 
of  the  wheel  4  feet,  what  power  must  be  applied  to  the  wheel 
to  raise  960  pounds  at  the  axle  ?  Ans.  1201bs. 

8.  If  the  diameter  of  the  axle  be  6  inches,  and  the  diameter 
of  the  wheel  4  feet,  what  power  must  be  applied  to  the  axle  to 
raise  120  pounds  at  the  wheel  ?  Ans.  9601bs. 

9.  If  the  diameter  of  the  axle  be  6  inches,  and  120  pounds 
applied  to  the  wheel  raise  960  pounds  at  the  axle,  what  is  the 
diameter  of  the  wheel  ?  Ans.  4  feet. 

10.  If  the  diameter  of  the  wheel  be  4  feet,  and  120  pounds 
applied  to  the  wheel  raise  960  pounds  at  the  axle,  what  is  the 
diameter  of  the  axle  ?  Ans.  6  inches. 


THE  PULLEY. 


The  pulley  is  a  small  wheel,  movable 
about  its  axis  by  means  of  a  cord,  which 
passes  over  it. 

When  the  axis  of  a  pulley  is  fixed, 
the  pulley  only  changes  the  direction  of 
the  power  ;  if  movable  pulleys  are 
used,  an  equilibrium  is  produced  when 
the  power  is  to  the  weight  as  one  to 
the  number  of  ropes  applied  to  them. 
If  each  movable  pulley  has  its  own 
rope,  each  pulley  will  be  double  the 
power. 


To  find  the  weight  that  may  be  raised  by  a  given  power. 

RULE.  — Multiply  the  power  by  the  number  of  cords  that  support  the 
weight,  and  the  product  is  the  weight. 

11.  What  power  must  be«  applied  to  a  rope,  that  passes  over 
one  movable  pulley,  to  balance  a  weight  of  400  pounds  ? 

Ans.  2001bs. 

12.  What  weight  will  be  balanced  by  a  power  of  10  pounds, 
attached  to  a  cord  that  passes  over  3  movable  pulleys  ? 

Ans.  601bs. 
29 


338  MECHANICAL   POWERS.  [SECT.  LXXX. 

13.  What  weight  will  be  balanced  by  a  power  of  144  pounds, 
attached  to  a  cord  that  passes  over  2  movable  pulleys  ? 

Ans.  576lbs. 

14.  If  a  cord,  that  passes  over  two  movable  pulleys,  be  at- 
tached to  an  axle  6  inches  in  diameter,  whose  wheel  is  60  inches 
in  diameter,  what  weight  may  be  raised  by  the  pulley,  by  ap- 
plying 144  pounds  to  the  wheel  ?  Ans.  57601bs. 

THE  INCLINED  PLANE. 

An  inclined  plane  is  a  plane  which  makes  an  acute  angle 
with  the  horizon. 

The  motion  t)f  a  body  descending  an  inclined  plane  is  uni- 
formly accelerated. 

The  force  with  which  a  body  descends  an  inclined  plane, 
by  the  force  of  attraction,  is  to  that  with  which  it  would  descend 
freely,  as  the  elevation  of  the  plane  to  its  length ;  or,  as  the 
size  of  its  angle  of  inclination  to  radius. 

To  find  the  power  that  will  draw  a  weight  up  an  inclined 
plane. 

RULE  —  Multiply  the  weight  by  the  perpendicular  Iteight  of  the  plane, 
and  divide  this  product  by  t/te  length. 

15.  An  inclined   plane  is  50  feet  in  length,  and  10  feet  in 
perpendicular  height;  what  power  is  sufficient  to  draw  up  a 
weight  of  1000  pounds  ?  Ans.  2001bs. 

16.  What  weight,  applied  to  a  cord  passing  over  a  single  pul- 
ley at  the  elevated  part  of  an  inclined  plane,  will  be  able  to 
sustain  a  weight  of  1728  pounds,  provided  the  plane  was  600 
feet  long,  and  its  perpendicular  height  5  feet  ?     Ans.  14f  Ibs. 

17.  A  certain  railroad,  one  mile  in  length,  has  a  perpendic- 
ular elevation  of  50ft. ;  what  power  is  sufficient  to  draw  up  this 
elevation  a  train  of  cars  weighing  20,0001bs.  ?     Ans.  189^-§. 

18.  An  inclined  plane  is  300  feet  in  length,  and  30  feet  in 
perpendicular  height ;  what  power  is  sufficient  to  draw  up  a 
weight  of  2000  pounds  ?  Ans.  200lbs. 

19.  An  inclined  plane  is  1000  feet  in  length,  and  100  feet  in 
perpendicular  height ;  what  power  is  sufficient  to  draw  up  this 
elevation  a  weight  of  5000  pounds  ?  Ans.  SOOlbs. 

20.  What  weight  applied  to  and  passing  over  a  single  pulley, 
at  the  elevated  part  of  an  inclined  plane,  will  be  able  to  sustain 
a  weight  of  70001bs.,  provided  the  plane  is  300  feet  long  and 
its  perpendicular  height  30  feet  ?  Ans.  7001bs. 


SECT.  LXIX.] 


MECHANICAL   POWERS. 


339 


THE  SCREW. 

The  screw  is  a  cylinder,  which  has 
either  a  prominent  part  or  a  hollow 
line  passmg  round  it  in  a  spiral  form, 
so  inserted  in  one  of  the  opposite 
kind  that  it  may  be  raised  or  de- 
pressed at  pleasure,  with  the  weight 
upon  its  upper,  or  suspended  beneath 
its  lower  surface. 

In  the  screw  the  equilibrium  will 
be  produced,  when  the  power  is  to 
the  weight  as  the  distance  between 
the  two  contiguous  threads,  in  a  di- 
rection parallel  to  the  axis  of  the 
screw,  to  the  circumference  of  the  circle  described  by  the 
power  in  one  revolution. 

To  find  the  power  that  should  be  applied  to  raise  a  given 
weight. 

RULE.  —  As  the  distance  between  the  threads  of  the  screiv  is  to  the  cir- 
cumference of  the  circle  described  by  the  power,  so  is  the  power  to  the 
weight  to  be  raised. 

NOTE.  —  One  third  of  the  power  is  lost  in  overcoming  friction. 

21.  If  the  threads  of  a  screw  be  1  inch  apart,  and  a  power 
of  100  pounds  be  applied  to  the  end  of  a  lever  10  feet  long, 
what  force  will  be  exerted  at  the  end  of  the  screw  ? 

Ans.  75,398.20+lbs. 

22.  If  the  threads  of  a  screw  be  £  an  inch  apart,  what  pow- 
er must  be  applied  to  the  end  of  a  lever  100  inches  in  length 
to  raise  100,000  pounds  ?  Ans.  79.5774-^-lbs. 

23.  If  the  threads  of  a  screw  be  £  an  inch  apart,  and  a  pow- 
er of  79.5774-j-  pounds  be  applied  to  the  end  of  a  lever  100 
inches  in  length,  what  weight  will  be  raised  ? 

Ans.  100,0001bs. 

24.  If  a  power  of  79.5774-)-  pounds  be  applied  to  the  end 
of  a  lever  100  inches  long,  and  by  this  force  a  weight  of 
100,000  pounds  be  raised,  what  is  the  distance  between  the 
threads  of  the  screw  ?  Ans.  £  an  inch. 

25.  If  a  power  of  79.5774+  pounds  be  applied  to  the  end 
of  a  lever,  raising  by  this  force  a  weight  of  100,000  pounds, 
what  must  be  the  length  of  the   lever,  if  the  threads  of  the 
screw  be  £  an  inch  apart  ?  Ans.  100  inches. 


340  SPECIFIC   GRAVITY.  [SECT.  LXXXI. 

WEDGE. 

The  wedge  is  composed  of  two  inclined  planes,  whose  bases 
are  joined. 

When  the  resisting  forces  and  the  power  which  acts  on  the 
wedge  are  in  equilibrio,  the  weight  will  be  to  the  power  as  the 
height  of  the  wedge  to  a  line  drawn  from  the  middle  of  the 
base  to  one  side,  and  parallel  to  the  direction  in  which  the  re- 
sisting force  acts  on  that  side. 

To  find  the  force  of  the  wedge. 

RULE.  —  As  half  tlie  breadth  or  thickness  of  the  head  of  the  wedge  is 
to  &ne  of  its  slanting  sides,  so  is  the  power  which  acts  against  its  head 
to  the  force  produced  at  its  side. 

26.  Suppose    100   pounds  to  be  applied  to  the  head  of  a 
wedge  that  is  2  inches  broad,  and  whose  slant  is  20  inches 
long,  what  force  would  be  affected  on  each  side  ? 

Ans.  20001bs. 

27.  If  the  slant  side  of  a  wedge  be  12  inches  long,  and  its 
head  1£  inches  broad,  and  a  screw  whose  threads  are  £  of  an 
inch  asunder  be  applied  to  the  head  of  this  wedge,  with  a  pow- 
er of  200  pounds  at  the  end  of  tlie  lever,  16  feet  long,  what 
would  be  the  force  exerted  on  the  sides  of  the  wedge  ? 

Ans.  5147184.2-f-lbs. 


SECTION  LXXXI. 
SPECIFIC    GRAVITY.* 

To  find  the  specific  gravity  of  a  body. 

RULE.  —  Weigh  the  body  both  in  water  and  out  of  water,  and 
note  the  difference,  which  will  be  the  weight  lost  in  water;  then,  as  the 
weight  lost  in  water  is  to  the  whole  weight,  so  is  the  specific  gravity  of 
water  to  the  specific  gravity  of  the  body.  But  if  the  body  whose  specific 
gravity  is  required  is  lighter  than  water,  affix  to  it  another  body  ht-nrii  r 
than  water,  so  that  the  mass  compounded  of  the  two  may  sink  together. 
Weigh  the  dense  body  and  the  compound  mass  separately,  both  in  water 
and  out  of  it  ;  then  find  how  much  each  loses  in  water  by  subtracting  its 
weight  in  water  from  its  weight  in  air ;  and  subtract  tlie  less  of  these 
remainders  from  the  greater ;  then  say,  as  the  last  remainder  is  to  the 

*  The  specific  gravity  of  a  body  is  its  weight  compared  with  water ; 
the  water  being  considered  1000. 


SECT.  LXXXII.]        STRENGTH  OF   MATERIALS.  341 

weight  of  the  body  in  air,  so  is  the  specific  gravity  of  water  to  the  spe- 
cific gravity  of  the  body. 

NOTE.  —  A  cubic  foot  of  water  weighs  1000  ounces. 

1.  A  stone  weighed  10  pounds,  but  in  water  only  6£  pounds. 
Required  the  specific  gravity.  Ans.  2608.6-J-. 

2.  Suppose  a  piece  of  elm  weigh  15  pounds  in  air,  and  that 
a  piece  of   copper,  which  weighs   18  pounds  in  air  and   16 
pounds  in  water,  is  affixed  to  it,  and  that  the  compound  weighs 
6  pounds  in  water.     Required  the  specific  gravity  of  the  elm. 

Ans.  600. 


SECTION  LXXXII. 
STRENGTH   OF   MATERIALS. 

THE  force  with  which  a  solid  body  resists  an  effort  to  sepa- 
rate its  particles  or  destroy  their  aggregation  can  only  become 
known  by  experiment. 

There  are  four  different  ways  in  which  the  strength  of  a  solid 
body  may  be  exerted ;  first,  by  resisting  a  longitudinal  tension ; 
secondly  by  its  resisting  a  force  tending  to  break  the  body  by  a 
transverse  strain ;  thirdly,  in  resisting  compression,  or  a  force 
tending  to  crush  the  body ;  and,  fourthly,  in  resisting  a  force 
tending  to  wrench  it  asunder  by  torsion.  We  shall,  however, 
only  consider  the  strength  of  materials  as  affected  by  a  trans- 
verse strain. 

When  a  body  suffers  a  transverse  strain,  the  mechanical  ac- 
tion which  takes  place  among  the  particles  is  of  a  complicated 
nature.  The  resistance  of  a  beam  to  a  transverse  strain  is  in 
a  compound  ratio  of  the  strength  of  the  individual  fibres,  the 
area  of  the  cross  section,  the  distance  of  the  centre  of  gravity 
of  the  cross  section  from  the  points  round  which  the  beam 
turns  in  breaking. 

The  following  are  the  facts  and  principles  on  which  mechan- 
ics make  their  calculations. 

1.  A  stick  of  oak  one  inch  square  and  twelve  inches  long, 
when  both  ends  are  supported  in  a  horizontal  position,  will  sus- 
tain a  weight  of  600  pounds ;  and  a  bar  of  iron  of  the  same 
dimensions  will  sustain  2190  pounds. 

2.  The  strength  of  similar  beams  varies  inversely  as  their 

29* 


342  STRENGTH   OF   MATERIALS.        [SECT.  LXXXII. 

lengths ;  that  is,  if  a  beam  10  feet  long  will  support  1000  pounds, 
a  similar  beam  20  feet  long  would  support  only  500  pounds. 

3.  The  strength  of  beams  of  the  same  length  and  depth  is 
directly  as  their  width ;  that  is,  if  there  be  two  beams,  each  20 
feet  long  and  6  inches  deep,  and  one  of  them  is  6  inches  wide 
and  the  other  but  3  inches,  the  former  will  support  twice  the 
weight  of  the  latter. 

4.  The  strength  of  beams  of  the  same  length  and  width  is  as 
the  squares  of  their  depths ;  that  is,  if  there  be  two  beams,  each 
of  which  is  20  feet  long  and  4  inches  wide,  but  one  is  6  inches 
deep  and  the  other  is  3  inches  deep,  their  strength  is  as  the 
squares  of -these  numbers.      Thus,  6x6  =  36;   3x3  =  9; 
that  is,  the  strength  of  the  former  is  to  the  latter  as  36  to  9.     It 
will,  therefore,  sustain  four   times  the  weight  of  the  latter. 
Thus,  36  -r-  9  =  4. 

5.  To  compare  the  strength  of  two  beams  of  the  same  length, 
but  of  different  breadth  and  depth,  we  multiply  their  widths  by 
the  squares  of  their  depths,  and  their  products  show  their  com- 
parative strength.     Thus,  if  we  wish  to  ascertain   how  much 
stronger  is  a  joist  that  is  2  inches  wide  and  8  inches  deep, 
than  one  of  the  same  length  that  is  4  inches  square,  we  mul- 
tiply 2  by  the  square  of  8,  and  4  by  the  square  of  4 ;  thus,  2 
X  8  X  8  =  128;   4  X  4  X  4  =  64  ;    128  H-  64  =  2.     Thus, 
we  see  that  although  the  quantity  of  material  in  one  joist  is  the 
same  as  in  the  other,  yet  the  former  will  sustain  twice  the 
weight  of  the  latter.     Hence  "  deep  joists  "  are  much  stronger 
than  square  ones,  which  have  the  same  area  of  a  transverse 
section. 

6.  To  compare  the  strength  of  two  beams  of  different  lengths, 
widths,  and  depths,  we  multiply  their  widths  by  the  squares  of 
their  depths,  and  divide  their  products  by  their  lengths,  and  their 
quotients  will  show  their  comparative  strength.     Therefore,  if 
we  wish  to  ascertain  how  much  stronger  is  a  beam  that  is  20 
feet  long,  8  inches  wide,  and  10  inches  deep,  than  one  10  feet 
long,  6  inches  wide,  and  5  inches  deep,  we  adopt  the  following 
formulas  :  — 

8X10X10  6X5X5__ 

~20~  ~IO~~ 

The  strength  of  the  former,  therefore,  is  to  the  latter  as  40  to 
15  ;  that  is,  if  the  first  beam  would  sustain  a  weight  of  40cwt., 
the  latter  would  sustain  only  15cwt. 

7.  Having  all  the  dimensions  of  one  beam  given,  to  find 


SECT.  LXXXII.]       STRENGTH  OF  MATERIALS.  343 

another,  part  of  whose  dimensions  are  known,  that  will  sustain 
the  same  weight.  We  multiply  the  width  of  the  given  beam 
by  the  square  of  its  depth,  and  divide  this  product  by  the  length, 
and  the  result  we  call  the  reserved  quotient ;  then,  if  we  have  the 
length  and  breadth  of  the  required  beam  given  to  find  the  depth, 
we  multiply  the  reserved  quotient  by  the  length  of  the  required 
beam,  and  divide  the  product  by  its  width,  and  the  quotient  is 
the  square  of  the  depth  of  the  required  beam.  If  the  length 
and  depth  of  the  required  beam  were  given  to  find  the  width, 
we  multiply  the  reserved  quotient  by  the  length  of  the  required 
beam,  and  divide  this  product  by  the  square  of  the  depth  of  the 
required  beam,  and  the  quotient  is  the  breadth.  But  if  the  width 
and  depth  of  the  required  beam  were  given  to  find  the  length, 
we  multiply  the  width  of  the  required  beam  by  the  square  of 
the  depth,  and  divide  this  product  by  the  reserved  quotient,  and 
the  result  is  the  length  of  the  required  beam. 

8.  A  triangular  beam  will  sustain  twice  the  weight  with  its 
edge  up  that  it  will  with  its  edge  down.     Hence  split-rails  hav e 
twice  the  strength  with  the  narrow  part  upward,  which  they 
have  with  the  narrow  part  downward. 

9.  In  making  the  above  calculations,  we  have  not  noticed  the 
weight  of  the  beam  itself,  and  in  short  distances  it  is  of  but  little 
consequence ;  but  where  a  long  beam  is  required,  its  weight  is 
of  importance  in  the  calculation. 

10.  A  beam  supported  at  one  end  will  sustain  only  one  fourth 
part  the  weight  which  it  would  if  supported  at  both  ends. 

11.  The  tendency  to  produce  fracture  in  a  beam  by  the  ap- 
plication of  a  weight  is  greatest  in  the  centre,  and  decreases 
towards  the  points  of  support ;  and  this  ratio  varies  as  .the  square 
of  half  the  length  of  the  beam  to  the  product  of  any  two  parts 
where  the  weight  may  be  applied.     Hence  the  tendency  of  a 
weight  to  break  a  bar  8  feet  long,  when  applied  to  the  centre, 
to  that  of  the  same  weight,  when  applied  3  feet  from  one  end, 
is  as  4  X  4  =  16  to  3  X  5  =  15. 

QUESTIONS  TO  BE  PERFORMED  BY  THE  PRECEDING  RULES. 

1.  If  a  stick  of  oak  1  inch  square  and  12  inches  long,  when 
both  ends  are  supported  in  a  horizontal  position,  will  sustain  a 
weight  of  600  pounds,  how  many  pounds  would  a  similar  stick 
sustain,  that  was  36  inches  long  ?  Ans.  200  pounds. 

2.  If  a  beam  4  inches  square  and  12  feet  long  would  support 
a  weight  of  1000  pounds,  how  many  pounds  would  a  similar 
beam  support,  that  was  3  feet  long  ?          Ans.  4000  pounds. 


344  STRENGTH   OF   MATERIALS.       [SECT.  LXXXII. 

3.  If  a  beam  20ft.  long,  4in.  wide,  and  6in.  deep,  will  sustain 
a  weight  of  2000lbs.,  how  many  pounds  would  a  similar  beam 
sustain,  that  was  Sin.  wide  ?  Ans.  15001bs. 

4.  If  a  beam  10  feet  long,  4  inches  wide,  and  3  inches  deep, 
will  sustain  a  weight  of  1000  pounds,  how  many  pounds  would 
a  similar  beam  support,  that  was  6  inches  deep  ? 

Ans.  4000  pounds. 

5.  If  a  beam  10  feet  long,  2  inches  wide,  and  4  inches  deep, 
will  sustain  a  weight  of  1000  pounds,  how  many  pounds  would 
a  beam,  that  is  10  feet  long,  4  inches  wide,  and  6  inches  deep, 
sustain  ?  Ans.  4500  pounds. 

6.  If  a  beam  2  feet  long,  2  inches  wide,  and  3  inches  deep, 
will  sustain  a  weight  of  4000  pounds,  what  will  a  beam,  that  is 
4  feet  long,  3  inches  wide^  and  6  inches  deep,  sustain  ? 

Ans.  12000  pounds. 

7.  If  a  beam  that  is  10  feet  long,  4  inches  wide,  and  6  inches 
deep,  will  sustain  a  weight  of  4  tons,  what  weight  will  a  beam 
sustain,  that  is  20  feet  long,  8  inches  wide,  and  10  inches  deep  ? 

Ans.  1 1£  tons. 

8.  If  a  beam  6  inches  square  and  8  feet  long  will   support  a 
weight  of  2000  pounds,  what  weight  will  a  beam  10  feet  long 
and  10  inches  square  sustain  ?  Ans.  7407£|  pounds. 

9.  If  a  beam  15  feet  long  and  5  inches  square  will  sustain  a 
weight  of  1200  pounds,  required  the  length  of  a  beam,  that  is 
8  inches  square,  that  will  sustain  a  weight  of  2000  pounds. 

Ans.  36  ff  f  feet. 

10.  If  a  beam  8  feet  long  and  7  inches  square  will  sustain 
a  weight  of  3000  pounds,  how  many  inches  square  must  be  the 
beam,  that  is  6  feet  long,  that  will  sustain  2000  pounds  ? 

Ans.  5.5-f-  inches. 

11.  If  a  bar  of  iron  10  feet  long,  2  inches  wide,  and  3  inches 
deep,  will  sustain  10  tons,  what  must  be  the  depth  of  a  bar,  that 
is  12  feet  long  and  3  inches  wide,  that  will  sustain  30  tons  ? 

Ans.  4.64-f-  inches. 

12.  If  it  require  1000  pounds  to  break  a  certain  beam,  that 
is  24  feet  long,  when  placed  in  its  centre,  required  the  weight 
necessary  to  break  it,  when  placed  within  4  feet  of  one  end  of 
the  beam?  Ans. 

13.  If  a  beam  6  inches  square  and  10  feet  long  will  sustain  a 
weight  of  2000  pounds  from  its  centre,  what  weight  would  a 
beam  of  the  same  material  sustain  that  is  10  inches  square  and 
12  feet  long,  if  the  weight  were  suspended  2  feet  from  the 
centre  ?  Ans. 


SECT.  Lxxxiii.j      ASTRONOMICAL  PROBLEMS.  345 

SECTION  LXXXIII. 
ASTRONOMICAL   PROBLEMS. 

PROBLEM    I. 

To  find  the  dominical  letter  for  any  year  in  the  present  cen- 
tury, and  also  to  find  on  what  day  of  the  week  January  will 
begin. 

RULE.  —  To  the  given  year  add  its  fourth  part,  rejecting  the  frac- 
tions ;  divide  this  sum  by  7 ;  if  nothing  remains,  the  dominical  letter  is 
A ;  but  if  there  be  a  remainder,  subtract  it  from  8,  and  the  residue  will 
show  the  dominical  letter,  reckoning  0  =  A,  2  =  B,  3=  C,  4  =  1), 
5  =E,  6  =  F,7=G.  These  letters  will  also  show  on  what  day  of  the 
week  January  begins.  For  when  A  is  the  dominical  letter,  January 
begins  on  the  Sabbath;  when  B  is  the  dominical  letter,  January  begins 
on  Saturday ;  C  begins  it  on  Friday ;  D,  on  Thursday;  E,  on  Wed- 
nesday; F,  on  Tuesday ;  G,  on  Monday. 

NOTE.  —  If  it  be  required  to  find  the  dominical  letter  for  the  last  centu- 
ry, proceed  as  above  ;  only,  if  there  be  a  remainder  after  division,  sub- 
tract it  from  7,  and  the  remainder  shows  the  dominical  letter,  reckoning 
1=A,  2  =  B,3=C,  4  =  D,  5=E,  6=F,  0  =  G. 

1.  Required  the  dominical  letter  for  1835. 

OPERATION. 

4)  1835         8  —  4  :=  4  z=  D  =i  dominical  letter. 

— -  As  D  is  the  dominical  letter,  January  began 

7)229*  on  Thursday,  and   the  fourth  day  was   the 

327-4       Sabbath. 

2.  Required  the  dominical  letter  for  1836. 

OPERATION. 

4)  1836       8  —  6  =  2  =  B  and  C  —  dominical  letters. 

459 
7)2295  I*1  leaP  years  there  are  two  dominical  letters. 

327  -6    ^e  ^ast  letter>  ^»  *s  f°r  Januar7  and  February, 
and  B  for  the  remainder  of  the  year.     As  C  is 
the  dominical  letter,  January  began  on  Friday,  and  the  third 
day  was  the  Sabbath. 

3.  Required  the  dominical  letter  for  1841  ?  Ans.  C. 

4.  Required  the  dominical  letter  for  1899.  Ans.  A. 

5.  Required  the  dominical  letters  for  1896.     Ans.  D  and  E. 

6.  What  is  the  dominical  letter  for  1786  ?  Ans.  A. 

7.  What  is  the  dominical  letter  for  1837  ?  Ans.  A. 


346  ASTRONOMICAL  PROBLEMS.       [SECT.  LXXXIII. 

PROBLEM    II. 

To  find  on  what  day  of  the  week  any  given  day  of  the  month 
will  happen. 

RULE.  —  Find  by  the  last  problem  the  dominical  letter  for  the  given 
year,  and  on  what  clay  in  January  will  be  the  first  Sabbath ;  and  the 
corresponding  day  in  the  succeeding  months  will  be  as  follows  :  —  Wed- 
nesday for  February  ;  Wednesday  for  March  ;  Saturday  for  April; 
Monday  for  May ;  Thursday  for  June ;  Saturday  for  July ;  Tuesday 
for  August ;  Friday  for  September ;  Sabbath  for  October;  Wednesday 
for  November  ;  Friday  for  December.  Having  found  the  day  of  the 
week  for  any  day  in  the  month,  any  other  day  may  be  easily  obtained,  as 
may  be  seen  in  the  following  example. 

8.  Let  it  be  required  to  ascertain  on  what  day  of  the  week 
was  the  25th  day  of  September,  1838. 

The  dominical  letter  for  1838  is  G  ;  therefore,  the  7th  of 
January  was  the  Sabbath,  and,  by  the  above  rule,  the  7th  of 
February  was  Wednesday,  the  7th  of  March  was  Wednesday, 
the  7th  of  April  was  Saturday,  the  7th  of  May  was  Monday,  the 
7th  of  June  was  Thursday,  the  7th  of  July  was  Saturday,  the 
7th  of  August  was  Tuesday,  the  7th  of  September  was  Friday. 
If  the  7th  was  Friday,  the  14th,  the  21st,  and  the  28th  were 
Fridays.  And  if  the  21st  was  Friday,  the  22d  was  Saturday, 
the  23d  was  the  Sabbath,  the  24th  was  Monday,  and  the  25th, 
the  day  required,  was  Tuesday. 

The  following  distich  will  assist  the  memory  in  finding  the 
corresponding  days  of  the  month  :  — 

At  Dover  Dwells  George  Brown  Esquire, 
Good  Christian  Friend,  And  David  Friar. 

It  will  be  recollected,  that  the  initial  A  is  for  the  Sabbath,  B 
for  Monday,  C  for  Tuesday,  D  for  Wednesday,  E  for  Thurs- 
day, F  for  Friday,  and  G  for  Saturday. 

NOTE. — When  it  is  leap  year,  the  days  of  the  week,  after  February, 
will  be  one  day  later  than  on  other  years. 

9.  Required  the  day  of  the  week  for  the  4th  of  July,  1836. 
We  find  the  dominical  letters  to  be  B  and  C,  the  3d  day  of 

January  therefore  was  on  the  Sabbath,  and  by  the  above  rule 
the  3d  day  of  July  would  have  been  on  Saturday  ;  but  as  it  was 
leap  year,  the  3d  of  July  was  one  day  later ;  that  is,  it  was 
on  the  Sabbath  ;  and,  if  the  3d  was  the  Sabbath,  the  4th  of  July 
was  on  Monday. 

10.  On  what  day  of  the  week  will  be  Dec.  8,  1849  ? 

Ans.  Saturday. 


SECT.  LXXXIV.J     MISCELLANEOUS  QUESTIONS.  347 

11.  On  what  day  will  happen  July  4, 1857  ?  Ans.  Saturday. 

12.  On  what  day  will  March  begin  in  the  year  1890  ? 

Ans.  Saturday. 

13.  On  what  day  of  the  week  was  our  Independence  de- 
clared ?  Ans.  Thursday. 

14.  There  will  be  a  "  Transit  of  Venus,"  December  8, 1874 ; 
on  what  day  of  the  week  will  it  happen  ?         Ans.  Tuesday. 

15.  On  what  day  of  the  week  were  you  born  ? 


SECTION  LXXXIV. 
MISCELLANEOUS  QUESTIONS. 

1.  What  number  must  7f  be  multiplied  by,  that  the  product 
may  be  6f  ?  Ans.  |f . 

2.  What  number  is  that,  which,  being  multiplied  by  half  it- 
self, the  product  shall  be  4£  ?  Ans.  3. 

3.  What  fraction  is  that,  which,  being  divided  by  llf ,  the 
quotient  shall  be  5  ?  Ans. 

4.  What  part  of  7f  is  9f-  ?  Ans.  f  f- f-  or 

7 

5.  Reduce to  a  simple  fraction.  Ans.  T5T. 

19f 

6.  Add  f-  of  a  ton  to  •&  of  a  cwt.      Ans.  12cwt.  Iqr.  8f  Ib. 

7.  If  the  earth  make  one  complete  revolution  in  23  hours 
56  minutes  3  seconds,  in  what  time  does  it  move  one  degree  ? 

Ans.  3m.  59"  20'". 

8.  Multiply  f-  of  9^  by  |  of  |-  of  8}.  Ans.  449f . 

¥  "B"  T 

9.  Divide  12"^  of  -f-  of  100  by  A  of  7f 

TD"  »*  Ans.  363j|f§f . 

10.  What  number  is  that  to  which  if  f  of  f-  be  added  the 
sum  will  be  1  ?  Ans.  %%. 

11.  A  certain  gentleman,  at  the  time  of  his  marriage,  agreed 
to  give  his  wife  §  of  his  estate,  if,  at  the  time  of  his  death,  he 
left  only  a  daughter,  and,  if  he  left  only  a  son,  she  should  have 
£  of  his  property ;  but,  as  it  happened,  he  left  a  son  and  a 
daughter,  by  which  the  widow  received  in  equity  $  2400  less 
than  if  there  had  been  only  a  daughter.   What  would  have  been 
his  wife's  dowry  if  he  had  left  only  a  son  ?          Ans.  $  2100. 

12.  A  gentleman  being  asked  what  o'clock  it  was,  said  that 


348  MISCELLANEOUS  QUESTIONS.      [SECT.  LXXXIV. 

it  was  between  5  and  6  ;  but,  to  be  more  particular,  he  said  that 
the  minute  hand  had  passed  as  far  beyond  the  6  as  the  hour  hand 
wanted  of  being  to  the  6  ;  that  is,  that  the  hour  and  minute  hands 
made  equal  acute  angles  with  a  line  passing  from  the  12  through 
the  6.  Required  the  time  of  day.  Ans.  32m.  18T67sec.  past  5. 

13.  A,  B,  and  C  are  to  share  $  100,000  in  the  proportion  of 
£,  £,  and  -^,  respectively ;  but  C's  part  being  lost  by  his  death, 
it  is  required  to  divide  the  whole  sum,  properly,  between  the 
other  two.       Ans.  A's  part  is  $  57,142f ,  and  B's  $  42,857f 

14.  A  father  devised  T7¥  of  his  estate  to  one  of  his  sons,  and 
T7F  of  the  residue  to  the  other,  and  the  remainder  to  his  wife. 
The  difference  of  his  sons1  legacies  was  found  to  be  257<£.  3s. 
4d.     What  money  did  he  leave  for  his  widow  ? 

Ans.  635<£.  Os.  10f$d. 

15.  In  the  walls  of  Balbec,  in  Turkey,  the  ancient  Heliop- 
olis,  there  are  three  stones  laid  end  to  end,  now  in  sight,  that 
measure  61  yards  in  length;  one  of  which  is  63  feet  long,  12 
feet  thick,  and  12  feet  broad  ;  what  is  its  weight,  supposing  its 
specific  gravity  to  be  3  times  that  of  water  ?     Ans.  759£  tons. 

16.  A  burden  of  2001bs.,  suspended  on  a  pole  4ft.  in  length, 
the  point  of  suspension  being  6in.  from  the  middle,  is  carried  by 
two  men,  the  ends  of  the  pole  resting  on  their  shoulders ;  how  much 
of  this  load  is  borne  by  each  man  ?      Ans.  1251bs.  and  751bs. 

17.  The  new  court-house  in  Boston  has  8  pillars  of  granite, 
each  25ft.  4in.  in  length,  4ft.  5in.  in  diameter  at  one  end,  and 
3ft.  5in.  in  diameter  at  the  other  end.     How  many  cubic  feet  do 
they  contain,  and  what  is  their  weight,  allowing  a  cubic  foot  to 
weigh  3000  ounces  ?        Ans.  2455.03  cubic  feet,  205.4  tons. 

18.  A  father,  dying,  left  his  son  a  legacy,  £  of  which  he 
spent  in  8  months ;  $-  of  the  remainder  lasted  him  12  months 
longer;    after  which  he  had  only  $410  left.     What  did  his 
father  bequeathe  him  ?  Ans.  $  956.66§. 

19.  A  butcher,  wishing  to  buy  some  sheep,  asked  the  owner 
how  much  he  must  give  him  for  20 ;  on  hearing  his  price,  he 
said  it  was  too  much ;  the  owner  replied,  that  he  should  have 
10,  provided  he  would  give  him  a  cent  for  each  different  choice 
of  10  in  20,  to  which  he  agreed.     How  much  did  he  pay  for 
the  10  sheep,  according  to  the  bargain  ?         Ans.  $  1847.56. 

20.  A  merchant  sold  goods  to  a  certain  amount,  on  a  com- 
mission of  4  per  cent.,  and,  having  remitted  the  net  proceeds 
to  the  owner,  he  received  £  per  cent,  for  prompt  payment, 
which  amounted  to  $  15.60.    What  was  the  amount  of  his  com- 
mission  ?  Ans.  $  260.00. 


SECT.  LXXXIV.]      MISCELLANEOUS  QUESTIONS.  349 

21.  A,  of  Boston,  remits  to  B,  of  New  York,  a  bill  of  ex- 
change on  London,  the  avails  of  which  he  wishes  to  be  invested 
in  goods  on  his  account.     B  having  disposed  of  the  bill  at  7^ 
per  cent,  advance,  he  received  $  9675.00,  and  having  reserved 
for  himself  £  per  cent,  on  the  sale  of  the  bill,  and  2  per  cent, 
for  commission,  what  will  remain  for  investment,  and  for  how 
much  was  the  bill  drawn  ? 

Ans.  For  investment,  $9461.58^;  the  bill  was  ,£2025. 

22.  Bunker  Hill  Monument  is  30ft.  square  at  its  base,  and  15ft. 
square  at  its  top  ;  its  height  is  220ft.     From  the  bottom  to  the 
top,  through  its  centre,  is  a  conical  avenue  15ft.  in  diameter  at 
the  bottom  and  about  lift,  at  the  top.     How  many  cubic  feet 
are  there  in  the  monument  ?  Ans.  86,068.518-f-ft. 

23.  A  hired  a  house  for  one  year  for  $  300 ;  at  the  end  of  4 
months  he  takes  in  M  as  a  partner,  and  at  the  end  of  8  months 
he  takes  in  P.    At  the  end  of  the  year  what  rent  must  each  pay  ? 

Ans.  A  pays  $  183£ ;  M  pays  $  83£ ;  P  pays  $  33^. 

24.  A  merchant  receives  on  commission  three  kinds  of  flour ; 
from  A  he  receives  20  barrels,  from  P  25  barrels,  and  from  C  40 
barrels.     He  finds  that  A's  flour  is  10  per  cent,  better  than  B's, 
and  that  B's  is  20  per  cent,  better  than  C's.    He  sells  the  whole 
at  $  6  per  barrel.     What  in  justice  should  each  man  receive  ? 

Ans.  A  receives  $  139£f|;  B,  $  158££f ;  C,  8211£Jf. 

25.  Bought  100  barrels  of  flour,  at  $  5  per  barrel,  and  im- 
mediately sold  it  on  a  credit  of  6  months.     The  note  which 
I  received  for  pay,  I  got  discounted  at  the  Suffolk  Bank,  and, 
on  examining  my  money,  I  found  that  I  had  gained  20  per  cent, 
on  my  purchase.     What  did  I  receive  per  barrel  for  the  flour  ? 

Ans.  $6.18!f|f. 

26.  Purchased  for  a  cloak  5^  yards  of  broadcloth,  that  was 
1|-  yards  wide ;  to  line  this,  I  purchased  flannel  that  was  £ 
yard  wide,  but  on  being  wet  it  shrunk  1  nail  in  width,  and  1 
yard  in  every  20  yards  in  length.     How  many  yards  of  flannel 
was  it  necessary  to  buy  ?  Ans.  12|f  f-  yards. 

27.  How  many  bricks  would  it  require  to  build  the  walls  of 
a  house  40  feet  long,  30  feet  wide,  and  20  feet  high,  admitting 
the  walls  to  be  a  foot  thick,  and  that  each  brick  was  8  inches 
long,  4  inches  wide,  and  2  inches  thick  ?     Ans.  73,440  bricks. 

28.  How  many  feet  of  boards  would  it  require  to  cover  a 
house,  that  was  40  feet  square,  and  whose  height  to  the  top  of 
the  plate  was  20  feet,  the  roof  projecting  1  foot  over  the  plate, 
and  coming  to  a  point  over  the  centre  of  the  house,  15  feet  above 
the  garret  floor  ?  Ans.  5367.7+  feet. 

30 


350  MISCELLANEOUS   QUESTIONS.      [SECT.  LXXXIV. 

29.  Lent  a  friend  $  700,  which  he  kept  20  months.     Some 
years  after  I  borrowed  of  him  £  300  ;  how  long  should  I  keep 
it  to  balance  the  favor  ?  Ans.  46§  months. 

30.  John  Lee  gave  half  of  his  estate  to  his  wife,  £  of  the 
remainder  to  his  oldest  son,  and  ^  of  the  residue  to  his  oldest 
daughter,  and  £  of  what  then  remained  to  be  distributed  to  his 
other  children,  who  received  $  200  each ;  required  the  number 
of  his  children,  and  the  value  of  his  estate. 

Ans.  12  children  ;  estate,  $  10,000. 

31.  A  and  B  set  out  to  travel  round  a  certain  island,  which 
is  80  miles  in  circumference.     A  travels  5  miles  a  day,  and  B 
7  miles  a  day.     How  far  must  B  travel  to  overtake  A  ? 

Ans.  448  miles. 

32.  If  24.4  cubic  inches  of  lead  weigh  16  pounds,  required 
the  number  of  feet  of  lead   pipe  that  can  be  made  'from  80 
pounds  of  lead,  the  diameter  of  the  pipe  to  be  1  inch,  and  the 
thickness  of  it  £  of  an  inch.  Ans.  123.24-f-  feet. 

33.  How  long  a  tube  can  be  made  from  a  cylinder  of  lead  8 
inches  long  and  2  inches  in  diameter,  and  through  the  centre 
of  which  is  a  hole  £  of  an  inch  in  diameter ;  the  tube  or  pipe 
to  be  £  of  an  inch  in  diameter,  and  f  of  an  inch  in  thickness  ? 

Ans.  39. 11+ feet. 

34.  Four  men,  A,  B,  C,  and  D,  bought  a  stack  of  hay,  con- 
taining 8  tons,  for  $  100.     A  is  to  have  12  per  cent,  more  of 
the  hay  than  B,  and  B  is  to  have  10  per  cent,  more  than  C,  and 
C  is  to  have  8  per  cent,  more  than  D.     Each  man  is  to  pay  in 
proportion  to  the  quantity  he  receives.     The  stack  is  20  feet 
high,  and  12  feet  wide  at  its  base,  it  being  an  exact  pyramid  ;  and 
it  is  agreed  that  A  shall  take  his  share  first  from  the  top  of  the 
stack,  B  is  to  take  his  share  the  next,  and  then  C  and  D.     How 
many  feet  of  the  perpendicular  height  of  the  stack  shall  each 
take,  and  what  sum  shall  each  pay  ? 

Ans.  A  takes  13.22+ft.,  and  pays  $  28.93£ff f|- ;  B  takes 
3.14-fft.,  and  pays  $25.83£lff£;  C  takes  1.94+ft.  and  pays 
$23.48£fm ;  D  takes  1.68-fft.,  and  pays  S21.74£|£f£. 

35.  Suppose  the  rails  of  a  railroad  to  be  5  feet  4  inches 
apart  at  the  place  of  the  wheels  bearing,  and  on  a  curve  line  of 
1200  feet  radius  for  the  outer  rail.     Suppose  the  wheels  of  the 
car  running  on  the  rails  to  be  firmly  fixed  to  the  axle,  and  that 
it  is  5  feet  from  the  outside  of  the  flange  of  one  wheel  to  the 
outside  of  the  flange  of  the  opposite  wheel ;  that  from  the  outer 
side  of  one  wheel  to  the  outer  side  of  its  opposite  wheel  is  5 
feet  8  inches  ;  that  the  diameter  of  each  wheel  at  the  outside 


SECT.  LXXXIV.]      MISCELLANEOUS   QUESTIONS.  351 

of  the  flange  is  3  feet ;  the  face  of  the  wheels  to  be  so  bevelled 
that  at  the  outside  of  each  wheel  the  diameter  of  the  wheel  is 
2  feet  11.5  inches,  and  that  the  axle  will  always  be  in  a  posi- 
tion square  across  the  two  rails.  In  what  part,  between  the 
two  wheels,  must  the  centre  of  gravity  of  the  load  be  placed,  so 
that  the  weight  of  the  load  shall  bear  equally  on  each  rail  ? 

OPERATION. 

5ft.  4in.  —  5ft.  rr  4in. ;  and  4in.  -J-  2  =  2in.  =  the  place,  be- 
yond the  flange,  on  the  face  of  the  wheel,  where  the  wheel  bears 
on  the  rail.  The  perpendicular  width  of  the  face  of  the  wheel 
is  4  inches,  and  in  that  distance  the  wheel  diminishes  in  diam- 
eter 3ft.  — 2ft.  11. Sin.  =  .5in.  Then,  as  4in.  :  .5in.  :  :  2in.  : 
.25in. ;  and  so  the  diameter  of  the  wheel  at  the  bearing  is 
3ft.  —  .25in.  =  2ft.  1 1.75in.  As  the  radius  of  the  curve  of  the 
outer  rail  is  to  that  of  the  inner  rail,  so  is  the  diameter  of  the 
outer  wheel  at  the  place  of  bearing  to  that  of  the  inner  wheel ; 
viz.  as  1200ft. :  1200ft.  —  5ft.  4in.  =  1194ft.  8in.  : :  2ft.  11.75in. 
:  2ft.  11.59iin.  Then  the  difference  of  the  diameters  of  the 
two  wheels  at  the  places  of  bearing  is  35.75in.  —  35.591  = 
.158in.  Then,  to  see  how  much  on  the  perpendicular  face  of 
the  wheel  this  difference  in  diameter  will  vary  the  bearing ;  as 
3ft.  — 2ft.  11.5in.  =  .5in.  :  4in.  :  :  .158in.  :  1.27iin.  But  this 
difference  must  be  applied  half  to  each  wheel,  viz.  1.27  lin.  -5- 
2=.635in.,  which  brings  the  place  of  bearing  .635in.  nearer 
to  the  flange  on  the  outer  wheel,  and  .635in.  further  from  the 
flange  on  the  inner  wheel.  And  the  middle  point  between  the 
two  bearings  will  be  .635in.  from  the  centre  of  the  axle  towards 
the  inner  curve  ;  and  in  that  place  must  the  centre  of  gravity 
of  the  whole  load  be  placed,  to  make  the  weight  of  the  load 
bear  equally  on  each  rail ;  viz.  .635in.  from  the  middle  of  the 
axle  towards  the  inner  rail,  Answer. 

36.  The  dimensions  of  a  bushel  measure  are    18£  inches 
wide,  and  8  inches  deep ;  what  should  be  the  dimensions  of  a 
similar  measure  that  would  contain  8  bushels  ? 

Ans.  37in.  wide,  16in.  deep. 

37.  What  is  the  weight  of  a  hollow  spherical  iron  shell  5in. 
in  diameter,  the  thickness  of  the  metal  being  lin.,  and  a  cubic 
inch  of  iron  weighing  ^§§  of  a  pound  ?        Ans.  13.23871bs. 

38.  At  a  certain  time  between  2  and  3  o'clock,  the  minute 
hand  was  between  3  and  4.     Within  an  hour  after,  the  hour 


352  MISCELLANEOUS  QUESTIONS.      [SECT.  LXXXIT. 

hand  and  minute  hand  had  exactly  changed  places  with  each 
other.  What  was  the  precise  time  when  the  hands  were  in 
the  first  position  ?  Ans.  2h.  15m.  56-^sec. 

39.  Required  the  contents  of  a  cube,  that  will  contain  a  globe 
20  inches  in  diameter ;  also  of  a  cube,  that  may  be  inscribed  in 
said  globe. 

Ans.  Larger  cube  8000  cub.  in. ;  smaller,  1539  cub.  in. 

40.  If  in  a  pair  of  scales  a  body  weigh  90  pounds  in  one 
scale,  and  only  40  pounds  in  the  other,  required  the  true  weight, 
and  the  proportion  of  the  lengths  of  the  two  arms  of  the  balance- 
beam  on  each  side  of  the  point  of  suspension. 

Ans.  the  weight  601bs.,  and  the  proportions  3  to  2. 

41.  In  turning  a  one-horse  chaise  within  a  ring  of  a  certain 
diameter,  it  was  observed  that  the  outer  wheel  made  two  turns, 
while  the  inner  wheel  made  but  one  ;  the  wheels  were  each  4 
feet  high ;  and  supposing  them  fixed  at  the  distance  of  5  feet 
asunder  on  the  axletree,  what  was  the  circumference  of  the 
track  described  by  the  outer  wheel  ?  Ans.  62.83-f-  feet. 

42.  The  ball  on  the  top  of  St.  Paul's  church  is  6  feet  in  di- 
ameter.    What  did  the  gilding  of  it  cost,  at  3£d.  per  square 
inch?  Ans.  237£.  10s.  Id. 

43.  There  is  a  conical  glass,  6  inches  high,  5  inches  wide  at 
the  top,  and  which  is  £  part  filled  with  water.     What  must  be 
the  diameter  of  a  ball,  let  fall  into  the  water,  that  shall  be  im- 
mersed by  it  ?  Ans.  2.445-|-in. 

44.  A  certain  lady,  the  mother  of  three  daughters,  had  a 
farm  of  500  acres,  in  a  circular  form,  with  her  dwelling-house 
in  the  centre.     Being  desirous  of  having  her  daughters  settled 
near  her,  she  gave  to  them  three  equal  parcels,  as  large  as  could 
be  made  in  three  equal  circles,  included  within  the  periphery 
of  her  farm,  one  to  each,  with  a  dwelling-house  in  the  centre 
of  each ;  that  is,  there  were  to  be  three  equal  circles,  as  large 
as  could   be    made  within  a  circle  that  contained  500  acres. 
How  many  acres  did  the  farm  of  each  daughter  contain,  how 
many  acres    did  the  mother  retain,  how  far  apart  were  the 
dwelling-houses  of  the  daughters,  and  how  far  was  the  dwell- 
ing-house of  each  daughter  from  that  of  the  mother  ? 

Ans.  Each  daughter's  farm  contained  107  acres  2  roods 
31.22+  rods.  The  mother  retained  176  acres  3  roods  26.34-f- 
rods.  The  distance  from  one  daughter's  house  to  the  other 
was  148.119817-f-  rods.  The  mother's  dwelling-house  was 
distant  from  her  daughters'  85.51-|-  rods. 


SECT.  LXXXIV.]      MISCELLANEOUS  QUESTIONS.  353 

341 

45.  Required  the  cube  of  —  -  .  Ans.  ?8T. 

5*2f 

78— 

46.  Required  the  cube  root  of  ^-^3-  Ans.  §. 


fifi  2  ftQ  5 

47.  Multiply  the  cube  root  of         *T    by  the  square  of  ~. 

C5/2 

Ans.  Jfr. 


49.  Three  carpenters,  J.  Smith,  J.  Carleton,  and  John  Jones, 
agree  with  T.  Jenkins  to  build  him  a  house  and  find  the  mate- 
rials for  $  1000,  of  which  $  600  were  paid  in  advance,  and 
the   remainder   when   the   work  was   finished.     Carleton  and 
Jones  took  $  50  each  of  the  first  payment.     When  the  work 
was  completed,  it  appeared  by  Smith's  account,  who  received 
the  money  and  paid  the  bills,  for  which  he  was  allowed  a  com- 
pensation of  $  10,  that  he  had  paid  $  648.95,  exclusive  of  the 
payments  to  Carleton  and  Jones,  and  that  he  had  labored  63 
days.     Carleton  worked  51  days,  and  he  was  allowed  $  20  for 
the  use  of  his  shop,  &c.     Jones  worked  60  days,  and  his  bill 
for  boarding  the  men  they  hired  was  $  68.75.     Smith,  on  set- 
tling with  Jenkins  and  allowing  him  $23.15  charged  to  Carle- 
ton,  and  $  17.48  charged  to  Jones,  receives  the  balance  in  cash, 
and  on  exhibiting  his  statement  of  the  business  to  Carleton  and 
Jones,  he  pays  to  each  the  balance  due.     How  much  did  they 
make  per  day,  and  how  was  the  last  payment  disposed  of? 

Ans.   They  received  $  1.45  per  day,  and  Carteton  received 
$  20.80,  Jones  received  $  88.27,  and  Smith  received  8  250.30. 

50.  A  and  B  engaged  to  reap  a  field  for  90  shillings  ;  and 
as  A  could  reap  it  in  9  days,  they  promised  to  complete  it  in  5 
days.     They  found,  however,  that  they  were  obliged  to  call  in 
C,  an  inferior  workman,  to  assist  them  for  the  last  two  days,  in 
consequence  of  which  B  received  3s.  9d.  less  than  he  other- 
wise would  have  done.     In  what  time  could  B  and  C  each  reap 
the  field  ?     Ans.  B  could  reap  it  in  15  days,  and  C  in  18  days. 

51.  Samuel  Jenkins  and  Jarnes  Betton,  who  have  each  an 
apprentice,  engage   to   build   a   small   house    for  8  630.     By 
agreement  between  them,  Jenkins's  apprentice  is  to  be  allowed 
$'0.62£  per  day,  and  Betton's  $  1.00.     When  the  work  was 
finished,  it  appeared  that  Jenkins  had  worked  120  days,  and  his 
apprentice  100.     Betton  worked   96  days,  and  his  apprentice 

30* 


354  MISCELLANEOUS  QUESTIONS.      [SECT.  LXXXIV. 

135£  days.     While  doing  the  work,  they  received  each  $  210. 
What  is  each  man's  share  of  the  remaining  payment  ? 

Ans.  Due  to  Jenkins  $  92.50  ;  to  Betton  $  117.50. 

52.  A  merchant  tailor  bought  40  yards  of  broadcloth,  2£ 
yards  wide  ;  but  on  sponging  it,  it  shrunk  in  length  upon  every 
4  yards  half  a  quarter,  and  in  width,  one  nail  and  a  half  upon 
every  1£  yards.     To   line  this  cloth,  he  bought  flannel  5  quar- 
ters wide,  which,  being  wet,  shrunk  the  whole  width  on  every 
20   yards   in    length,  and    in    width    it    shrunk    half  a   nail. 
Required  the  number  of  yards  of  flannel  used  in  lining  the 
cloth.  Ans.  71T77j  yards. 

53.  What  is  the  square  of  .1  ? 

54.  If  a  stick  of  timber,  6  inches  square  and  10  feet  long, 
will  support  from  its  centre  1000  pounds,  how  many  pounds 
would  a  similar  stick,  that  is  10  inches  square  and  20  feet  long, 
support,  if  the  weight  were  suspended  4  feet  from  the  centre 
of  the  stick  ? 

55.  A  certain  gentleman  has  an  elegant  lamp,  to  which  are 
appended  72  brilliants,  each  of  which  had  occupied  a  particular 
place.     His  daughter,  one  day,  in   preparing  the  lamp,  dis- 
placed the  brilliants,  and  in  replacing  them,  found  that  she  had 
put  some  of  them  in  the  wrong  situation,  and  was  at  a  loss  to 
know  how  to  correct  the    mistake.      At  length  she  told  her 
father  that,  after  dinner,  she  would  begin  and  place  the  bril- 
liants in  all  the  situations  they  would  admit  of,  and  then  she 
would  be  sure  of  finding  the  correct  way  of  adjusting  them, 
and  that  she  would  not  take  her  tea  until  she  had  effected  it. 
Now,  supposing  she  were  to  place  them  in  all  the  various  ways 
they  would  admit,  how  long  would  she  be  obliged  to  wait  for 
her  tea,  provided  she  could  make  one  change  each  minute  ? 

Ans.  612344583768860868615240703852746727407780917 
8469732898382301496397838498722168927420416000000000 
0000000  minutes. 

In  order  that  the  pupil  may  have  some  general  idea  of  the 
time  denoted  by  the  answer  in  figures  to  the  above  question,  let 
him  suppose  a  globe  composed  of  fine  sand,  whose  circumfer- 
ence shall  be  equal  to  the  orbit  of  Herschel,each  cubic  inch  of 
sand  containing  one  million  of  particles  ;  and  then  suppose  the 
lady  at  the  end  of  every  million  of  years'  labor  to  remove  one 
of  these  particles,  and  she  will  have  removed  the  whole  globe, 
particle  by  particle,  millions  of  years  before  she  can  take  her  tea. 


APPENDIX.]  WEIGHTS  AND  MEASURES.  355 

APPENDIX. 
WEIGHTS  AND  MEASURES, 

WITH  AN  HISTORICAL  ACCOUNT  OF  STANDARDS. 

IN  commerce  and  in  science,  all  bodies  are  estimated  by 
number,  weight,  or  measure.  When  reckoned  by  number,  they 
are  referred  to  a  unit  as  a  standard  of  comparison,  and  when 
estimated  by  weight  or  measure,  there  is  always  a  reference  to 
some  certain  fixed  quantity,  as  a  pound,  a  gallon,  a  mile,  to 
which  the  quantity  measured  or  weighed  bears  a  specified  and 
definite  proportion. 

Weights  are  used  to  ascertain  the  gravity  of  bodies,  and 
measures  to  determine  their  magnitude,  or  the  space  which 
they  occupy. 

Standards  of  weights  or  measures  are  certain  quantities  of 
gravity  or  extension,  which  are  fixed  upon  as  those  with  which 
all  other  quantities  of  objects  reckoned  by  weight  or  measure 
are  to  be  compared  ;  and  such  standards  have  always  been  found 
necessary,  and  have  existed  in  every  age  and  nation.  It  has, 
however,  been  only  in  a  highly  civilized  state  of  society  that 
they  have  been  such  as  to  secure  an  accurate  and  equitable 
result  in  the  transaction  of  business ;  and  few  things  are  more 
indicative  of  a  cultivated  age  and  people,  than  the  exactness 
with  which  science  provides  and  adjusts  the  standards  of  com- 
parison, required  by  the  sale  and  interchange  of  commodities. 

In  the  early  stages  of  society,  the  ordinary  standards  of  weight 
and  measure  were  some  simple  objects  or  ideas,  with  which  all 
in  the  community  were  supposed  to  be  familiar.  Thus,  all 
measures  of  length  were  sometimes  reckoned  by  comparing 
them  with  the  human  foot ;  or,  for  the  sake  of  greater  definite- 
ness,  as  all  human  feet  were  not  of  the  same  length,  they  were 
referred  to  the  king's  foot  as  a  standard.  In  some  cases  the 
length  of  the  arm  was  used,  and  in  other  instances  the  length 
of  a  grain  or  corn  of  wheat  or  barley.  In  land  measure,  an 
acre  was  what  could  be  ploughed  by  a  yoke  of  oxen  in  a  day, 
traces  of  which  notion  we  have  in  the  Hebrew  and  Latin  words 
used  to  denote  an  acre,  which  properly  signify  a  yoke  or  pair, 
—  that  is,  a  yoke  or  pair  of  oxen. 

In  early  ages,  also,  weights  as  well  as  distances  were  meas- 


356  WEIGHTS  AND  MEASURES.  [APPENDIX. 

ured  by  grains  of  corn,*  arid  hence,  in  England  and  some  other 
European  countries,  the  lowest  denomination  of  weight  is  still  a 
grain.  Originally,  32  of  these  grains  were  reckoned  to  a  penny- 
weight, but  in  later  times  the  number  was  fixed  at  24. 

A  scruple  meant  originally  a  small  stone,  which  was  regard- 
ed as  equal  to  20  grains,  and  a  dram  (Greek  drachma)  was 
literally  what  one  could  hold  in  his  hand, —  the  word  being  de- 
rived from  a  verb  signifying  to  grasp  with  the  hand. 

With  standards  thus  variable  and  uncertain,  it  is  evident  that 
no  people  could  advance  far  in  commerce,  and  as  facilities  for 
commercial  intercourse  opened,  they  must  adopt  some  more 
satisfactory  methods  of  ascertaining  the  quantity  and  value  of 
what  they  bought  and  sold.  Some  fixed  and  permanent  stand- 
ards of  comparison  were  needed,  to  which  all  might  refer,  and 
in  which  all  should  have  confidence.  Accordingly,  kings  and 
legislators  early  gave  attention  to  this  subject,  and  endeavoured, 
not  only  to  provide  such  standards,  but  to  preserve  them  from 
alteration.  In  Rome,  the  model  weights  and  measures  were 
kept  for  safety  in  the  temple  of  Jupiter ;  and  among  the  Jews, 
they  were  preserved  first  in  the  Tabernacle,  and  afterwards  in 
the  Temple,  and  their  custody  committed  to  the  sacerdotal  fam- 
ily. In  England,  the  standard  yard,  to  which,  as  we  shall  see, 
all  the  legal  measures  and  even  weights  of  the  kingdom  are  ul- 
timately referred,  has  for  ages  been  kept  with  the  greatest  care 
by  the  government.  This  yard,t  as  history  informs  us,  was  in- 
troduced or  revived  as  a  standard  of  measure  by  Henry  I.,  who 
lived  at  the  beginning  of  the  12th  century,  and  who  ordered 
that  the  ulna  or  ancient  ell,  answering  to  the  modern  yard, 
should  be  of  the  exact  length  of  his  own  arm,  and  all  other 
measures  founded  upon  it.  This  is  the  historical  origin  of  the 
present  English  standard  of  measure,  and  it  is  said  to  have  been 
preserved  to  the  present  time  without  any  sensible  variation! 

In  1742,  the  Royal  Society  caused  a  yard  to  be  made  from 
a  very  careful  comparison  of  the  standard  ells  or  yards  of  the 
reigns  of  Henry  VII.  and  Elizabeth,  which  had  been  kept  at 
the  Exchequer.  In  1758,  a  committee  of  the  House  of  Com- 
mons recommended  that  a  rod  which  had  been  made  by  their 
order  from  that  of  the  Royal  Society,  and  marked  "  Standard 
Yard,  1758,"  should  be  declared  the  legal  standard  of  all  meas- 
ures of  length.  This  rod  consisted  of  a  solid  brass  bar  a  little 

*  Corn  in  the  English  sense,  meaning  wheat,  barley,  &c.,  and  not  our 
Indian  corn  or  maize. 

The  word  yard  meant  originally  a  rod  or  shoot,  and  cannot  be  sup- 
posed to  have  had  any  very  definite  length. 


APPENDIX.]  WEIGHTS  AND  MEASURES.  357 

more  than  an  inch  square,  and  39.06  inches  long.  At  about 
an  inch  and  a  half  from  each  end,  there  was  inserted  a  gold  pin 
or  stud,  and  in  these  pins,  at  the  distance  of  36  inches  from 
each  other,  are  two  points,  which  are  intended  to  designate  the 
exact  length  of  the  yard.  The  subject  was  further  considered 
by  another  committee,  and  a  second  rod  made  by  the  same 
man,  and  as  an  exact  copy  of  the  abovef  which  is  known  as  the 
"  Standard  Yard  of  1760,"  and  which  was  declared  by  Parlia- 
ment in  the  "  Act  of  Uniformity,"  in  1824,  which  took  effect 
January  1,  1826,  to  be  the  legal  standard  of  measure  for  Great 
Britain.  As  the  distance  between  the  two  fixed  points  will 
vary  with  the  temperature  of  the  rod,  the  Act  provides  that  it 
shall  be  at  the  temperature  of  62°  Fahrenheit. 

This  standard  yard,  which  was  to  be  denominated  the  "  Im- 
perial Yard,"  was  intended  to  afford  a  fixed  standard  of  meas- 
ure, which  could  only  be  lost  with  the  destruction  or  mutilation 
of  the  standard  rod.  But  this,  it  was  foreseen,  might  easily 
occur,  as  in  fact  it  did  happen  in  1834,  at  the  burning  of  the 
two  Houses  of  Parliament,  when  the  imperial  model  shared  the 
fate  of  the  other  valuable  relics  which  were  consumed  in  that 
ancient  pile  ;  and  to  provide  against  such  an  accident,  it  was 
declared  that  the  imperial  or  standard  yard,  as  compared  with 
a  pendulum  vibrating  seconds  in  the  latitude  of  London,  should 
bear  the  proportion  of  36  to  39.1393  inches.  To  procure  fur- 
ther accuracy,  this  pendulum  was  to  move  in  a  vacuum,  at  the 
level  of  the  sea,  and  at  the  temperature  of  62°  Fahrenheit. 
And  thus  was  fixed  an  absolute  and  invariable  standard  of  linear 
measure,  by  which,  according  to  the  Act  above  named,  all 
measures  of  extension  are  determined,  whether  the  same  be 
lineal,  superficial,  or  solid.  A  third  part  of  this  standard  yard 
is  a  legal  foot,  a  twelfth  part  of  the  foot  a  legal  inch,  5£  such 
yards  a  pole,  220  a  furlong,  and  1760  a  mile.  A  legal  rood 
of  land  contains  1210  square  yards,  and  an  acre  4840  square 
yards,  or  160  square  poles. 

By  the  same  Act,  all  measures  of  capacity  are  determined 
by  the  imperial  gallon,  which  contains  277.274  cubic  inches. 
Two  such  gallons  make  a  peck,  one  fourth  of  a  gallon  a  quart, 
&c.  As  this  standard  gallon  is  determined  by  cubic  inches, 
we  see  that  it  is  ultimately  referred  to  the  standard  yard  as 
the  basis  on  which  it  is  founded.  The  Act  provides,  however, 
that  it  may  also  be  determined  by  weight,  in  which  case  the 
measure  to  contain  a  gallon  must  be  of  a  capacity  to  hold  10 
pounds,  avoirdupois  weight,  of  distilled  water,  weighed  in  air, 


358  WEIGHTS  AND  MEASURES.  [APPENDIX. 

at  the  temperature  of  62°  Fahrenheit,  the  barometer  being 
at  30  inches. 

For  a  standard  of  weights,  the  law  of  England  makes  the 
pound  Troy*  to  contain  5760  grains, —  one  cubic  inch  of  dis- 
tilled water  weighed  as  above,  weighing  252.458  such  grains, 
—  and  the  pound  Avoirdupois  to  contain  7000  such  grains. 

The  Act  of  Parliament  which  brought  the  standards  of  Eng- 
lish weights  and  measures  to  their  present  state  of  comparative 
perfection  was  passed  in  1835.  This  Act  prohibits  the  use  of 
the  well-known  Winchester  bushel,  and  also  of  heaped  measure. 
The  subject  of  weights  and  measures  is  still  under  consideration 
by  scientific  men  in  England,  and  further  improvements  have1 
been  proposed,  and  will  probably  be  introduced.  It  has  even 
been  suggested  by  commissioners  appointed  by  government,  that 
the  system  of  coinage  be  changed,  and  one  having  decimal  pro- 
portions adopted. 

The  system  of  weights  and  measures  established  by  law  in 
the  United  States  is  essentially  the  same  as  the  English ;  but 
there  is  not,  even  at  this  day,  that  uniformity  of  practice  in  this 
country,  which  the  interests  of  extensive  trade  require.  At  the 
organization  of  the  Federal  Government,  authority  was  given 
to  Congress  to  regulate  this  important  matter,  but  no  laws  have 
as  yet  been  enacted  by  that  body  to  secure  a  uniform  system 
through  all  the  States.  By  an  order  of  Congress,  in  June,  1836, 
a  set  of  standard  weights  and  measures  was  prepared  for  the 
use  of  each  custom-house,  and  for  each  State,  and  these  consti- 
tute the  legal  standards  of  the  nation,  so  far  as  any  national 
standard  can  be  said  to  exist.  The  standards  thus  provided 
were  similar  to  those  used  in  England  prior  to  the  Act  of  1824, 
and  they  lack  that  accuracy  and  scientific  character  which  are 
exceedingly  desirable  in  a  country  so  extensive  and  so  commer- 
cial as  the  United  States.  Many  of  the  States  of  the  Union 
have  attempted  to  reduce  their  weights  and  measures  to  a  uni- 
form system,  and  not  a  little  legislation  has  been  directed  to  this 
end  ;  but  generally  with  very  little  effect.  And  until  Congress 
shall  take  up  the  matter  in  good  earnest,  and  with  the  aid  of  sci- 
entific men,  but  little  can  be  done  to  secure  a  result  so  important. 

™hh  t°Tn  °f  ?°  T^t  Tr°7and  Avoirdupois  has  been  variously 
given  but  the  most  probable  explanation  is  this :  that  the  former  is  de- 
nved  from  avoirs  (avend),  the  ancient  name  for  goods  or  chattels,  and 
poids  (weight.)  The  word  Troy  has  probably  no  reference  to  a  toin  in 
th?  m  '  I"  TS  former^uPP°fd,  hut,  as  applied  to  weight,  originated  in 
the  monkish  name  of  Troy  Novant,  which  was  given  to  London,  and 
founded  on  an  ancient  legend ;  so  that  Troy  weight  is  properly  London 


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